Mathematics Tricks & Tips of binomial for NDA
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Tricks & Tips of binomial for NDA

General Term and Middle Term and coefficient in a Binomial Expansion

`"General Term in a binomial expression :"`

`T_(r+1) = text()^(n)C_r x^(n-r) a^r`


`"The middle term in the expansion of"` ` (a + x)^n` is as follows:

`"Case I :"` lf `n` is even, then `(n/2 +1 )` th term is the middle term.

`"Case II :"` If `n` is odd then `( n+1 )/2`th term and `(n+3)/2` th term are the middle terms.
Q 2773291146

The expansion of `(x- y)^n , n >= 5` is done in the descending powers of `x`. If the sum of the fifth and sixth terms is zero, then `x/y` is equal to
NDA Paper 1 2017
(A)

`(n - 5)/6`

(B)

`(n - 4)/5`

(C)

`5/(n - 4)`

(D)

`6/(n - 5)`

Solution:

`T_5+T_6=0`

`text()^n C_4 x^(n-4) (-y)^4 + text()^n C_5 x^(n-5) (-y)^5=0`

`(text()^n C_4)/(text()^n C_5) x ^(n-4-n+5) (-y)^(-1) =-1`

`x/y =(text()^n C_5)/(text()^n C_4) =(n-4)/5`
Correct Answer is `=>` (B) `(n - 4)/5`
Q 2261101925

Consider the expansion of `(1 + x)^(2n+1)`
If the coefficients of `x^r` and `x^(r + 1)` are equal in the
expansion, then `r` is equal to
NDA Paper 1 2015
(A)

`n`

(B)

`(2n -1)/2`

(C)

`(2n + 1)/2`

(D)

`n + 1`

Solution:

We have, `(1 + x)^(2n+1)`

General term , `T_(r+1) = text()^(2n+1)C_r x^r`

We have, ` text()^(2n+1)C_r = text()^(2n+1)C_(r+1)`

`∵ 2n + 1 = r + r + 1 `

`=> 2r = 2n `

`=> r = n`
Correct Answer is `=>` (A) `n`
Q 2231112022

Consider the expansion of `(1 + x)^(2n+1)`

The average of the coefficients of the two middle terms
in the expansion is
NDA Paper 1 2015
(A)

` text()^(2n+1)C_(n+2)`

(B)

` text()^(2n+1)C_(n)`

(C)

` text()^(2n+1)C_(n - 1)`

(D)

` text()^(2n)C_(n+1)`

Solution:

Since, `2n + 1` is odd.

Hence`( 2n+1+1)/2` and `( 2n+1+3)/2` are two middle terms

i.e. `(n + 1)` th and `(n + 2)` th term are two middle terms.

`:. (text()^(2n + 1)C_n + text()^(2n + 1)C_(n+1) )/2 = (text()^(2n + 1 +1 )C_(n+1))/2`

` = 1/2 text()^(2n + 2)C_(n+1) = 1/2 . ( 2n + 2)/(n+1) . text()^(2n + 1)C_n = text()^(2n + 1)C_n`
Correct Answer is `=>` (B) ` text()^(2n+1)C_(n)`
Q 1722278131

In the expansion of ` ( x^3 - 1/x^2)^n`
where `n` is a positive integer, the sum of the coefficients of `x^5`
and `x^(10)` is `0`.

What is `n` equal to?
NDA Paper 1 2014
(A)

`5`

(B)

`10`

(C)

`15`

(D)

None of these

Solution:

In the expansion of `(x + a)^n`.

General term, `T_(r + 1) = text()^(n)C_r x^(n-r) . a^r`

`:.` In the expansion of `( x^3 - 1/x^2)^n`

General term, `T_(r + 1) = text()^(n)C_r (x^3)^( n- r) . (1/x^2 )^r`

` = text()^( n)C_r . x^(3n- 3r) . (-1)^r . r^(- 2r)`

` = text()^( n)C_r . (-1)^r . x^(3n- 5r)` ...........(1)

For the coefficient `x^5`

put `3n - 5r = 5`

`=> 5r = 3n - 5`

`:. r = (3n)/5 - 1`

`:.` Coefficient of `x^5 = text()^(n)C_((3n)/5 -1)^-1^((3n)/5 -1)`
Correct Answer is `=>` (C) `15`
Q 2353523444

In the expansion of ` ( x^3 - 1/x^2)^n` where `n` is a positive integer, the sum of the coefficients of `x^5` and `x^(10)` is `0`.

What is the sum of the coefficients of the two middle terms?
NDA Paper 1 2014
(A)

0

(B)

1

(C)

-1

(D)

None of these

Solution:

In the expansion of `(x + a)^n`.

General term, `T_(r + 1) = text()^(n)C_r x^(n-r) . a^r`

`:.` In the expansion of `( x^3 - 1/x^2)^n`

General term, `T_(r + 1) = text()^(n)C_r (x^3)^( n- r) . (1/x^2 )^r`

` = text()^( n)C_r . x^(3n- 3r) . (-1)^r . r^(- 2r)`

` = text()^( n)C_r . (-1)^r . x^(3n- 5r)` ...........(1)

Since, `n = 15`

:. Total term in the expansion of `(x^3 - 1/x^2)^(15)` is 16

So, middle term `= (n/2) th` term and `(n/2 +1) th` term

`= (16/2) th` term and `( 16/2 +1)`th term

`=8` th term and `9` th term

So ` (x^3 - 1/x^2)^15` has two middle terms.

`T_8 = T_(7+1) = text()^15 C_7 (-1) ^7 * x ^(3 xx 15 -5 xx 7)`

`= - text()^15 C_7 * xC^10;` from eq (1)

and `T_9 = T_(8+1) = text()^15 C_8 (-1) ^8 * x^(3 xx -5 xx 8)`

`=text()^15C_8 * x^5`

Now, the sum of the coefficients of the two middle terms

`= -text()^15C_7 + text()^15C_8`

`=- text()^15C_7 + text()^15C_7 \ \ \ \ \ \ ( because text()^nC_r = text()^n C_(n-1))`

`=0`
Correct Answer is `=>` (A) 0
Q 1619091810

Consider the expansion ` (x^2 + 1/x )^(15)`

Consider the following statements

1. There are `15` terms in the given expansion.
2. The coefficient of `x^(12)` is equal to that of `x^3`

Which of the above statement (s) is/are correct?
NDA Paper 1 2014
(A)

Only 1

(B)

Only 2

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

1. We know that, `(a+ b)^n` have total number of

terms is `n + 1`.

So, `( x^2 + 1/x)^(15)` have `16` terms.

Hence, Statement `1` is false.

2. For coefficient of `x^(12)` ,

`30- 3r = 12 => r = 6 => text()^(15)C_6`

and for coefficient of `x_3` ,

`30- 3r = 3 => r = 9 => text()^(15)C_9`

`:. text()^(15)C_6 = text()^(15)C_9`

Hence, Statement `2` is correct.
Correct Answer is `=>` (B) Only 2
Q 1619191910

Consider the expansion ` (x^2 + 1/x )^(15)`

What is the sum of the coefficients of the middle. terms
in the given expansion?
NDA Paper 1 2014
(A)

`C(15, 9)`

(B)

`C(16, 9)`

(C)

`C(16,8)`

(D)

None of these

Solution:

We have, ` (x^2 + 1/x )^(15)`

Since, `n` is odd.

So, it has two middle terms `T_8` and `T_9` .

`:. T_8 + T_9 = text()^(15)C_7 + text()^(15)C_7`

`= text()^(16)C_8 quad ( ∵ text()^(n)C_(r-1) + text()^(n)C_r = text()^(n+1)C_r)`
Correct Answer is `=>` (C) `C(16,8)`
Q 2319345219

How many terms are there in the expansion of
`(1 + 2x + x^2) ^(10)` ?
NDA Paper 1 2013
(A)

`11`

(B)

`20`

(C)

`21`

(D)

`30`

Solution:

Given, `(1 + 2x + x^2)^(10) = {(1 + x)^2}^(10) = (1 + x)^(20)`

`:.` Total term `= 20 + 1 = 21`
Correct Answer is `=>` (C) `21`
Q 2319356210

What is the middle term in the expansion of

` ( 1 - x/2)^8` ?
NDA Paper 1 2011
(A)

` (35x^4)/8`

(B)

` (17x^5)/8`

(C)

` (35x^5)/8`

(D)

None of these

Solution:

Total number of terms in `( 1 - x/2)^8 = 9`

`∵ n = 8` (even)

`:.` Middle term `= (n/2 + 1)` th term

So, the middle term is `5`th term.

Hence ` T_5 = T_( 4 + 1) = text()^8C_4 (1)^4 ( - x/2 )^4 = ( 8 xx 7 xx 6 xx 5 )/ ( 4 xx 3 xx 2 xx 1 ) xx x^4 /2^4`

` = (70x^4)/(16) = (35x^4)/8`
Correct Answer is `=>` (A) ` (35x^4)/8`
Q 2309356218

What is the coefficient of `x^(17)` in the expansion of
`(3x - x^3/6)^9` ?
NDA Paper 1 2010
(A)

`(189)/8`

(B)

`(567)/2`

(C)

`(21)/(16)`

(D)

None of these

Solution:

Let `T_(r + 1)` be the term which contains `x^(17)`

`:. T_(r+1) = text()^9C_r (3x)^(9-r) (- x^3/6)^r` .

`= text()^9C_r 3^(9 - r) (-1)^r x^(9 + 2r)/6^r`

`=>` Put `9 + 2r = 17` for the coefficient of `x^(17)`

` => 2r = 17- 9 => r = 8/2 = 4`

`:.` Required coefficient, `T_(4 +1) = text()^9C_4 3^5/6^4 = (126 xx 3)/(16) = (189)/8`
Correct Answer is `=>` (A) `(189)/8`
Q 2562091835

If `A` and `B` are coefficients of `x^n` in the
expansions of `(1 + x)^(2n)` and `(1 + x)^(2n - 1)`
respectively, then `A /B` is equal to
WBJEE 2011
(A)

`4`

(B)

`2`

(C)

`9`

(D)

`6`

Solution:

Given, `A=` Coefficient of `x^n` in `(1 + x)^(2n)`

and `B =` Coefficient of `x^n` in `(1 + x)^(2n -1)`

`:. A = text()^(2n)C_n ` and `B= text()^(2n-1)C_n`

`:. A/B = (text()^(2n)C_n)/(text()^(2n-1)C_n) = (2n)/n`

`=2`
Correct Answer is `=>` (B) `2`
Q 1550878714

The middle term in the expansion of `(x-1/x)^18` is:
BITSAT 2006
(A)

`-15/4`

(B)

`15/4`

(C)

`-9/4`

(D)

`4`

Solution:

The general term in the expansion `(x - 1/x)^18`is given by

`T_(r+1) = text()^18C_r(x)^(18-r) (-1/x)^r`

Here, `n= 18`

`therefore` The middle term is `T_(9+1)`, Where `r = 9`

`therefore T_(9+1) = text()^18C_9(-1)^9 x^(18-2r)`

`= -text()^18C_9 x^(18-18) = -text()^18C_9 = - 15/4`
Correct Answer is `=>` (A) `-15/4`
Q 2339556412

What is the coefficient of `x^4` in the expansion of
` ( ( 1- x)/(1 + x) )^2` ?
NDA Paper 1 2010
(A)

`-16`

(B)

`16`

(C)

`8`

(D)

`- 8`

Solution:

` ( ( 1- x)/(1 + x) )^2 = (1- x)^2 (1 + x)^(-2)`

`= (1 - 2x + x^2 ) (1 + x)^(-2)`

`= (1 - 2x + x^2 ) (1 - 2x + 3x^2 - 4x^3 + 5x^4 - ... )`

`:.` Coefficient of `x^4` in `((1- x)/(1 + x)) = 5 + 8 + 3 = 16`
Correct Answer is `=>` (B) `16`
Q 2359456314

What is the number of terms in the expansion of
`(a + b +c)^n`, where `n in N`?
NDA Paper 1 2010
(A)

`n + 1`

(B)

`n + 2`

(C)

`n(n + 1)`

(D)

`((n + 1)(n + 2))/2`

Solution:

Required number of terms in `(a+ b +c)^n`

` = text()^(n + 2)C_2 = ((n + 2) !)/(2! n!)`

`= ((n + 2)(n + 1) n!)/(2.n!) = ((n + 1)(n + 2))/2`
Correct Answer is `=>` (D) `((n + 1)(n + 2))/2`
Q 2580034817

The number of irrational terms in the
binomial expansion of `(3^(1/5) + 7^(1/3))^100` is
WBJEE 2015
(A)

`90`

(B)

`88`

(C)

`93`

(D)

`95`

Solution:

General term of `(3^(1/5) + 7^(1/3))^100` is given by

`T_(r+1) = text()^(100)C_r (3^(1/5))^(100-r) (7^(1/3) )^r`

`= text()^(100)C_r * 3^((100-r)/5) * 7^(r/3)`

For a rational term, `(100 -r)/5 ` and `r/3` must be integer.

Hence, `r=0, 15, 30, 45, 60, 75, 90`

`:.` There are seven rational terms.

Hence, number of irrational terms

`= 101-7=94`
Q 2414512459

The value of r for which the coefficients of
`(r- 5)th` and `(3r + 1)th` terms in the
expansion of `(1 + x)^12` are equal, is
UPSEE 2011
(A)

`4`

(B)

`9`

(C)

`12`

(D)

None of these

Solution:

Since, coefficient of `(r- 5) th` term `=` coefficient of ` (3r + 1) th` term

`text()^12 C_(r-6) =text()^12 C_(3r)`

`=> r-6 =3r`

or `12-r+6=3r`

`=> 2r =-6` or `4r=18`

`=> r=-3` or `r=18/4`

Hence, no value of `r` exist, because `r` neither
be negative nor in fraction.
Correct Answer is `=>` (D) None of these
Q 2436156072

The value of `x`, for which the `6th` term in the expansion

`{ 2^(log 2 sqrt ( (9^(x-1) +7) ) ) + 1/( 2^(1/5) log_2 (3^(x-1) +1) ) }^7` is `84`, is equal to
UPSEE 2013
(A)

`4`

(B)

`3`

(C)

`-2`

(D)

`1`

Solution:

Given, expression

`= [ sqrt(9^(x-1) +7) + 1/(3^(x-1) +1)^(1/5) ]^7`

`:. T_6 = T_(5+1)`

`= text()^(7)C_5 (sqrt ( 9^(x-1) + 7) )^(7-5) {1/(3^(x-1) +1)^(1/5) }^(5)`

`= 21 (9^(x-1) +7) * 1/(3^(x-1) +1)`

`=21 * ( 3^(2x -2) +7)/(3^(x-1) +1) = 84` (given)

`=> 3^(2x-2) +7 = 4 (3^(x-1) +1)`

`=> y^2 + 7 - 4y - 4 = 0`

where `y = 3^(x-1)`

`:. y =3`

`=> 3^(x-1) = 3`

`=> x-1 = 1`

`=> x =2` and `y=1`

`=> 3^(x-1) = 3^0`

`=> x-1 = 0 => x =1`
Correct Answer is `=>` (D) `1`
Q 2502691538

If `(1-x+x^2)^n = a_0+a_1x+.................+a_(2n) x^(2n)` then the value of `a_0+a_2+a_4+...........+a_(2n)` is
WBJEE 2010
(A)

`3^n+1/2`

(B)

`3^n-1/2`

(C)

`(3^n-1)/2`

(D)

`(3^n+1)/2`

Solution:

Given `(1-x+x^2)^n = a_0+a_1x+.............+a_(2n)x^(2n)`..........(i)

`x = 1 , ` then from Eq(i)

`1 = a_0+a_1+........a_(2n)` ..............(ii)

and If `x = -1,` then from Eq(i)


`3^n = a_0-a_1+a_2-a_3+.............+a_(2n)`.............(iii)


Adding Eqs (ii) and (iii)

`1+3^n = 2[a_0+a_2+a_4+..........a_(2n)]`


`=> (1+3^n)/2 = a_0+a_2+a_4+..........a_(2n)`
Correct Answer is `=>` (D) `(3^n+1)/2`
Q 2542491333

If in the expansion of `(a- 2b)^n`, the sum of the 5th and 6th term is zero, then the value of `a/b` is
WBJEE 2010
(A)

`(n-4)/5`

(B)

`(2(n-4))/5`

(C)

`5/(n-4)`

(D)

`5/(2(n-4))`

Solution:

We know `(a-2b)^n = underset(r = 1) overset(n) Sigma text()^nC_r (a)^(n-r) (-2b)^r`

the `(r+1) ` th term = `t_(r+1) = text()^nC_r (a)^(n-r) . (-2b)^r`



`therefore t_5+t_6 = 0`


`=> text()^nC_4 (a)^(n-4) (-2b)^4+text()^nC_5(a)^(n-5) (-2b)^5 = 0`



`=> (n!)/(4!(n-4)!) a^(n-4) (-2b)^4 = -(n!)/(5!(n-5)!) (a)^(n-5) (-2b)^5`


`=> 1/(n-4)xx a = (-1)/5(-2b)`


`=> a/b = (2(n-4))/5`
Correct Answer is `=>` (B) `(2(n-4))/5`
Q 2314056859

The middle term in the expansion of
`(1 + 1/x)^(10)` is :
BITSAT Mock
(A)

`text()^(10)C_5 1/x^5`

(B)

`-text()^(10)C_6 1/x^6`

(C)

`- text()^(10)C_5 1/x^5`

(D)

`text()^(10)C_6 1/x^6`

Solution:

The given expression is `(1 + 1/x)^(10)`

Here `n = 10` which is an even number

So, `((10 + 1)/2)^(th)` i.e. `6` is the middle term

`∴ T_6 = T_(5 + 1) = text()^(10)C_5 (1)^(10 − 5) (1/x)^5`

`[ ∵ T_(r + 1) = text()^nC_r (a)^(n − r) b^r` where `r = 5, n = 10, a = 1, b = 1 // x]`

`⇒ T_6 = text()^(10)C_5 1/x^5`
Correct Answer is `=>` (A) `text()^(10)C_5 1/x^5`
Q 2580356217

If `n` is even, then in the expansion of `(1 + x^2/(2!) + x^4/(4!) + ... )^2` , then the coefficient of `x^n` is
BCECE Stage 1 2013
(A)

` 2^n/(n!)`

(B)

`(2^n - 2)/(n!)`

(C)

`( 2^(n-1) - 1)/(n!)`

(D)

`(2^(n - 1))/(n!)`

Solution:

`(1 + x^2/(2!) + x^4/(4!) + ... )^2 = ( (e^x + e^(-x))/2)^2`

` = 1/4 ( e^(2x) + e^(-2x) + 2)`

` = 1/4 { 2( 1 + (2x)^2/(2!) + (2x)^4/(4!) + ... ) + 2 }`

So, coefficient of `x^n` (n even) `= 1/2 { 2^n/(n!) } = 2^(n -1)/(n!)`
Correct Answer is `=>` (D) `(2^(n - 1))/(n!)`

Greatest Term and Greatest Coefficients in the Expansion of `(x + a)^n`

Greatest Term :

- If `T_r` and `T_(r + 1)` are the rth and `(r + 1)`th terms in the expansion of `(1 + x )^n` , then

`(T_(r+1))/(T_r) = ( text()^(n)C_(r) x^r )/(text()^(n)C_(r-1) x^(r-1) ) = (n-r +1)/r x`


Greatest Coefficients in the Expansion of `(x + a)^n` :

(i) lf `n` is even, `text()^(n)C_(r)` is greatest when `r = n/2` i.e. greatest coefficient is ` text()^(n)C_(n/2)`.

(ii) If `n` is odd , `text()^(n)C_(r)` is greatest when `r = ( n-1 )/2` or `r = (n+1)/2`

i.e. greatest coefficients ` text()^(n) C_((n-1)/2)` and ` text()^(n) C_((n+1)/2)` .

Q 2815434369

The greatest term (numerically) in the expansion of `(3- 5x)^11` when `x = 1/5` is

(A)

`55 xx 3^9`

(B)

`46 xx 3^9`

(C)

`55 xx 3^6`

(D)

`46 xx 3^6`

Solution:

We have , `(3 -5x)^11 = 3^11 ( 1 - (5x)/3)^11`

`= 3^11 ( 1 - 1/3 )^11\ \ \ \ \ \ [ because x = 1/5 ]`

`:. m = (|x| ( n + 1) )/(| x| +1) \ \ \ \ \ \ [ because - 1/3 < 0 ]`

`= (( | - 1/3| ) ( 11 +1))/((| -1/3| ) +1 ) = 3`

The greatest terms in the expansion are `T_3` and `T_4`

`:.` Greatest term (When `r =2` )

`= 3^11 | T_ (2 +1) |`

`= 3^11 | text()^11 C_2 ( -1/3)^2 | = 55 xx 3 ^9`
Correct Answer is `=>` (A) `55 xx 3^9`
Q 2517523480

Find numerically the greatest term in the expansion of `(2 + 3x)^9`, when `x = 3 //2`.

Solution:

Let `T_(r+ 1)` be the greatest term in the expansion of `(2 + 3x)^9` , we

have

` T_(r + 1)/T_r = ((9 - r + 1)/r) | (3x)/2 | = ( (10 - r)/r) | 3/2 xx 3/2 | = ( 90 - 9r)/(4r)`

`:. T_(r + 1)/T_r >= 1`

` = ( 90 - 9r)/(4r) >= 1 => 90 >= 13r`

`:. r <= (90)/(13) = 6 (12)/(13)` or ` r <= 6 (12)/(13)`

`:.` Maximum value of `r` is `6`.

So, greatest term ` = T_(6 + 1) = text()^9C_6 (2)^(9-6) (3x)^6`

`= text()^9C_3 . 2^3 . ( 3 xx 3/2)^6 = ( 9.8.7)/(1.2.3) . (2^3 . 3^(12))/2^6 = ( 7 xx 3^(13))/2`
Q 1419756610

The largest term in the expansion of `((b/2) +(b/2))^100` is

(A)

`b^100`

(B)

`(b/2)^100`

(C)

`text()^100C_50 * (b/2)^100`

(D)

None of these

Solution:

If `T_(r+1) ge T_r`

`=> T_(r+1)/T_r ge 1`

`=> ((100-r+1)/r ) * (b/2)/(b/2) ge 1`

`=> 101 -r ge r`

`=> 2r le 101`
`:.` `r le 50.5`

`:.` `r=50`

`:.` Greatest term `T_(r+ 1) = T_(50+1)`

`= text()^100C_50 (b/2)^50 (b/2)^50`

`= text()^100C_50 (b/2)^100`
Correct Answer is `=>` (C) `text()^100C_50 * (b/2)^100`
Q 2510234119

If `text()^nC_(r - 1) = 10, text()^nC_r = 45` and `text()^nC_(r+1) = 120`, then
`r` equals to
BCECE Stage 1 2013
(A)

`1`

(B)

`2`

(C)

`3`

(D)

`4`

Solution:

Using ` (text()^nC_r)/(text()^nC_(r-1)) = (n - r + 1)/r`

` (text()^nC_r)/(text()^nC_(r-1)) = (45)/(10)` and `(text()^nC_(r+1))/ (text()^nC_r) = (120)/(45)`

` => (n - r + 1)/r = 9/2`

and `(n-r)/(r + 1) = 8/3`

` => 8/3 (r + 1 ) + 1 = 9/2 r`

` => 16r + 16 + 6 = 27 r`

` => 11r = 22`

` :. r = 2`
Correct Answer is `=>` (B) `2`

Finding the Independent term in binomial expression

Q 2200491318

In the expansion of `( sqrt(x) + 1/(3x^(2) ))^(10)` the value of constant
term (independent of `x`) is
NDA Paper 1 2015
(A)

`5`

(B)

`8`

(C)

`45`

(D)

`90`

Solution:

We have `( sqrt(x) + 1/(3x^(2) )^(10))`

`:. T_(r +1) = text ()^(10)C_(r) x^ ((10-r)/(2)) (3) ^(-r) = text ()^(10)C_(r) (3) ^(-r) x ^((10-5r)/(2))`

Since, the term independent of `x` is `(10-5r)/2 = 0`, then

`r = 2`

Put `r = 2`, we get

`T_(3) = text ()^(10)C_(2) (3)^(- 2) = (10 xx 9)/(1 xx 2) xx 1/9 = 5`
Correct Answer is `=>` (A) `5`
Q 2343423343

In the expansion of ` ( x^3 - 1/x^2)^n`
where `n` is a positive integer, the sum of the coefficients of `x^5`
and `x^(10)` is `0`.

What is the value of the independent term?
NDA Paper 1 2014
(A)

5005

(B)

7200

(C)

-5005

(D)

-7200

Solution:

In the expansion of `(x + a)^n`.

General term, `T_(r + 1) = text()^(n)C_r x^(n-r) . a^r`

`:.` In the expansion of `( x^3 - 1/x^2)^n`

General term, `T_(r + 1) = text()^(n)C_r (x^3)^( n- r) . (1/x^2 )^r`

` = text()^( n)C_r . x^(3n- 3r) . (-1)^r . r^(- 2r)`

` = text()^( n)C_r . (-1)^r . x^(3n- 5r)` ...........(1)



For the independent term ,

put `3n - 5r =0` from eq (1)

`=> 5r = 3n = 3 xx 15 ` `(because n =15)`

`=> 5r=3 xx 3 xx 5`
`:. r=9`

Now, put the value of r in Eq. (i), we get,

`T_(9 +1) = text()^15C_9 (-1) ^9 * x^(3xx 15 - 5 xx 9)`

`=> T_10 = -text()^15C_9 * x^0 = - text()^15 C_9`

`=> T_10 = 15 C_6` `(because text()^n C_r = text()^n C_(n -r) )`

`= -(15 !) /(6! 9!)` `{ because text()^nC_r = (n!)/(r!(n -r)!)}`

`= (15xx14xx13xx12xx11xx10)/(6xx5xx4xx3xx2xx1)`

`= -5005`

So, the value of the independent term is - 5005.
Correct Answer is `=>` (C) -5005
Q 1619891710

Consider the expansion ` (x^2 + 1/x )^(15)`

What is the independent term in the given expansion?
NDA Paper 1 2014
(A)

2103

(B)

3003

(C)

4503

(D)

None of these

Solution:

We have, ` (x^2 + 1/x )^(15)`

` T_( r + 1) = text()^(15)C_r (x^2)^(15-r) (1/x)^r`

` = text()^(15)C_r x^(30 - 2r - r) = text()^(15)C_r x^(30 - 3r)`

For independent term,

` 30 - 3r = 0 => r = 10`

Put `r = 10`, we get

` T_(10 + 1) = text()^(15)C_(10) = (15!)/(10! 5!)`

`=( 15 xx 14 xx 13 xx 12 xx 11 xx 10!)/( 10! xx 1 xx 2 xx 3 xx 4 xx 5) = 3003`
Correct Answer is `=>` (B) 3003
Q 1669891715

Consider the expansion ` (x^2 + 1/x )^(15)`

What is the ratio of coefficient of `x^(15)` to the term
independent of `x` in the given expansion?
NDA Paper 1 2014
(A)

`1`

(B)

`1/2`

(C)

`2/3`

(D)

`3/4`

Solution:

For coefficient of `x^(15)`

`30- 3r = 15 => r = 5`

So, the coefficient of `x^(15)` is `text()^(15)C_5`.

and coefficient of independent of `x` is

`30 - 3r = 0 => r = 10`

So, coefficient of independent of `x` is `text()^(15)C_(10)`.

`:.` Required ratio ` = (text()^(15)C_5)/(text()^(15)C_(10))`

` = (text()^(15)C_5)/(text()^(15)C_5)`

` = 1`
Correct Answer is `=>` (A) `1`
Q 2369845715

The value of the term independent of `x` in the
expansion of `(x^2 - 1/x)^9` is
NDA Paper 1 2012
(A)

`9`

(B)

`18`

(C)

`48`

(D)

`84`

Solution:

Given, `( x^2 - 1/x)^9`

General term, `T_(r + 1) = text()^nC_r (a)^(n - r) x^r `, in `(a+ x)^n`

Similarly, `T_(r +1) = text()^9C_r ,(x^2)^(9 - r) . ( - 1/x)^r` in `( x^2 - 1/x)^9`

`= text()^9C_r x^(18 - 2r) (-1)^r x^(-r)`

`= text()^9C_r x^(18 - 3r) (-1)^r` .....(i)

For independent term,

Put `18 - 3r = 0`

`=> 3r = 18`

` => r = 6`

`:. T_(6 + 1) = text()^9C_6 x^(18 - 18).(-1)^6`

`=> T_7 = text()^9C_6·1`

`= (9·8·7)/(3·2 ·1) = 84`
Correct Answer is `=>` (D) `84`
Q 2379156016

What is the ratio of coefficient of `x^(15)` to the term
independent of `x` in `(x^2 + 2/x)^(15)`?
NDA Paper 1 2011
(A)

`1//64`

(B)

`1//32`

(C)

`1//16`

(D)

`1//4`

Solution:

Binomial expression `= ( x^2 + 2/x)^(15)`

General term, `T_( r + 1) = text()^(15)C_r ,(x^2)^(15 - r) - ( 2/x)^r`

`= text()^(15)C_r x^(30 - 2r) (2)^r. x^(-r)`

`= text()^(15)C_r 2^r . x^(30 -3r)` ........(i)

For the coefficient of `x^(15)`,

Put `30 - 3r = 15`

`=> 3r = 15 => r = 5`

So, `T_(5+1) = text()^(15)C_5 2^5 ·x^(30 - 3 xx 5)`

`= text()^(15)C_5 2^5 . x^(15)`

Coefficient of `x^(15)` in `(x^2 + 2/x )^(15) = text()^(15)C_5 . 2^5` ........(ii)

For the term independent of `x,`

Put `30 - 3r = 0 => r = 10`

So, `T_(10 + 1) = text()^(15)C_(10) . 2^(10) .x^(30- 3 xx 10)`

`= text()^(15)C_(10) .2^(10) . x^0`

Coefficient of `x^0` ,i.e., independent of `x = text()^(15)C_(10) . 2^(10)` .......(iii)

`:.` Ratio of coefficient of `x^(15)` to the term independent of `x`,

` = (text()^(15)C_5 ·2^5)/ (text()^(15)C_(10) ·2^(10)) = (text()^(15)C_(10) ·2^5)/ (text()^(15)C_(10) ·2^(10))`

`= 1/2^5 = 1/(32)`
Correct Answer is `=>` (B) `1//32`
Q 2439678512

The term independent of `x` in `[sqrt(x/3)+ sqrt(3/(2x^2))]^10` is
BCECE Stage 1 2011
(A)

`text()^10 C_1`

(B)

`5//12`

(C)

`1`

(D)

None of these

Solution:

`T_(r+1) =text()^10 C_1 (sqrt (x/3))^(10-r) (sqrt(3/(2x^2)))^r`

`text()^10C_r (1/sqrt 3)^(10-r) (sqrt(3/2))^r x^(s-r/2-r)`

Let `T_(r + 1)` be the term independent of `x`.

`:. 5-r/2-r=0 => (3r)/2=5`

`=> r=10/3`

`:. r` is not a whole number.

`:.` Given expression does not have any term independent of `x`.
Correct Answer is `=>` (D) None of these
Q 1619891710

Consider the expansion ` (x^2 + 1/x )^(15)`

What is the independent term in the given expansion?
NDA Paper 1 2014
(A)

2103

(B)

3003

(C)

4503

(D)

None of these

Solution:

We have, ` (x^2 + 1/x )^(15)`

` T_( r + 1) = text()^(15)C_r (x^2)^(15-r) (1/x)^r`

` = text()^(15)C_r x^(30 - 2r - r) = text()^(15)C_r x^(30 - 3r)`

For independent term,

` 30 - 3r = 0 => r = 10`

Put `r = 10`, we get

` T_(10 + 1) = text()^(15)C_(10) = (15!)/(10! 5!)`

`=( 15 xx 14 xx 13 xx 12 xx 11 xx 10!)/( 10! xx 1 xx 2 xx 3 xx 4 xx 5) = 3003`
Correct Answer is `=>` (B) 3003
Q 2580101017

The term independent of x in the expansion of `(x- 1/x)^4 (x +1/x)^3` is
BCECE Stage 1 2014
(A)

-3

(B)

0

(C)

3

(D)

1

Solution:

`(x-1/x)^4 (x + 1/x)^3`

`= ( text()^4C_0 x^4 - text()^4C_1x^2 + text()^ 4C_2 - text()^4C_3 1/x^2 + text()^4C_4 1/x^4) xx (text()^3C_0 x^3 + text()^3C_1 x + text()^3C_2 1/x + text()^3 C_3 1/x^3 )`

It is clear that there is no term free from x on RHS.

Hence, the term independent of x is zero.
Correct Answer is `=>` (B) 0
Q 2343423343

In the expansion of ` ( x^3 - 1/x^2)^n`
where `n` is a positive integer, the sum of the coefficients of `x^5`
and `x^(10)` is `0`.

What is the value of the independent term?
NDA Paper 1 2014
(A)

5005

(B)

7200

(C)

-5005

(D)

-7200

Solution:

In the expansion of `(x + a)^n`.

General term, `T_(r + 1) = text()^(n)C_r x^(n-r) . a^r`

`:.` In the expansion of `( x^3 - 1/x^2)^n`

General term, `T_(r + 1) = text()^(n)C_r (x^3)^( n- r) . (1/x^2 )^r`

` = text()^( n)C_r . x^(3n- 3r) . (-1)^r . r^(- 2r)`

` = text()^( n)C_r . (-1)^r . x^(3n- 5r)` ...........(1)



For the independent term ,

put `3n - 5r =0` from eq (1)

`=> 5r = 3n = 3 xx 15 ` `(because n =15)`

`=> 5r=3 xx 3 xx 5`
`:. r=9`

Now, put the value of r in Eq. (i), we get,

`T_(9 +1) = text()^15C_9 (-1) ^9 * x^(3xx 15 - 5 xx 9)`

`=> T_10 = -text()^15C_9 * x^0 = - text()^15 C_9`

`=> T_10 = 15 C_6` `(because text()^n C_r = text()^n C_(n -r) )`

`= -(15 !) /(6! 9!)` `{ because text()^nC_r = (n!)/(r!(n -r)!)}`

`= (15xx14xx13xx12xx11xx10)/(6xx5xx4xx3xx2xx1)`

`= -5005`

So, the value of the independent term is - 5005.
Correct Answer is `=>` (C) -5005
Q 2865245165

If the independent term in the expansion of `(sqrtx + k /x^2)^10` is 405, then consider the following statements.

I. The third term is independent of x.
II. The value of k, is ± 3.
III. Total number of terms in the expansion is 10.

Which of the above statement(s) is/are correct?

(A)

I, II

(B)

II, III

(C)

I, III

(D)

I, II and Ill

Solution:

General term of the expansion

`(sqrtx + k /x^2 )^10` is

`T_(r +1) = text()^10C_r ( sqrtx)^(10 -r) ( k/x^2)^r`

`= text()^10 C_r ( x) ^((10-r)/2) ( k)^r * (x) ^(-2r)`

`= text()^10 C_r (k)^r (x) ^((10 -5r)/2)` ................(i)

Since, ( r + 1 )th term is independent of x.

`:. (10 -5r)/2 = 0 => r =2`

Thus, 3rd term is independent of x.

So, Statement I is correct.

Put r = 2 in Eq. (i), we get

`T_3 = text()^10 C_2 k^2 => 405 = 45k^2`

`=> k^2 = 9 => k = pm 3`

Thus, Statement II is correct. and total number of terms in `(sqrtx + k/x^2)^10 ` is 11

So, Statement III is not correct.
Correct Answer is `=>` (A) I, II

Properties of Binomial Coefficients

Q 2713180940

The value of `[C (7, 0) + C (7, 1)] + [C(7, 1) + C(7, 2)] + ... + [C(7, 6) + C(7, 7)]` is


NDA Paper 1 2017
(A)

`254`

(B)

`255`

(C)

`256`

(D)

`257`

Solution:

`=(text()^7 C_6+ text()^7 C_1) +( text()^7 C_1+ text()^7 C_2)+ ........(text()^7 C_6+text()^7 C_7)`

`= text()^8 C_1+ text()^8 C_2+...........+ text()^8 C_7`

`= text()^8 C_0+text()^8 C_1+text()^8 C_2+..........+text()^8 C_7+ text()^8 C_8- text()^8 C_0- text()^8 C_8`

`= 2^8 - 2= 254`
Correct Answer is `=>` (A) `254`
Q 2731356222

What is `text()^47 C_4 + text()^51 C_3 + sum _ (j =2 ) ^5 text ()^(52-j)C_3` equal to ?
NDA Paper 1 2016
(A)

`text()^52C_4`

(B)

`text()^51C_5`

(C)

`text()^53C_4`

(D)

`text()^52 C_5`

Solution:

`text()^47 C_4 + text()^51 C_3 + sum _ (j =2 ) ^5 text ()^(52-j)C_3`

we know `text()^n C_r + text()^n C_(r-1)= text()^(n+1) C_r`

`= text()^(47)C_4 + text()^(51) C_3+ (text()^(50) C_3+ text()^(49) C_3+ text()^(48) C_3+ text()^(47) C_3)`

`= text()^(51) C_3+ (text()^(50) C_3+ text()^(49) C_3+ text()^(48) C_4)`

`= text()^51 C_3 + text()^51 C_3 = text()^52 C_4`
Correct Answer is `=>` (A) `text()^52C_4`
Q 2201112028

The sum of the coefficients of all the terms in the
expansion is `(1 + x)^(2n+1)`

NDA Paper 1 2015
(A)

`2^(2n - 1)`

(B)

`4^(n - 1)`

(C)

`2.4^n`

(D)

None of these

Solution:

To find the sum of coefficient of all terms, put `x = 1` in

the given expression `(1 + x)^(2n+1)` we get

` 2^(2n+1) = 2.2 ^(2n) = 2.4^n`
Correct Answer is `=>` (C) `2.4^n`
Q 1669091815

Consider the expansion ` (x^2 + 1/x )^(15)`

Consider the following statements
1. The term containing `x^2` does not exist in the given
expansion.
2. The sum of the coefficients of all the terms in the given
expansion is `2^(15)`

Which of the above statements is/are correct?
NDA Paper 1 2014
(A)

Only 1

(B)

Only 2

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

1. For coefficient of `x^2`

`30 - 3r = 2 => r = (28)/3 , r notin N`

So, `x^2` does not exist in the expansion.

Hence, Statement `1` is correct.

2. Now, `( x^2 + 1/4 ) ^(15) = text()^( 15)C_0 (x^2)^(15) + text()^( 15)C_1 (x^2)^(14) (1/x)`

` + ...+ text()^( 15)C_(15) (1/x)^(15)`

Put `x = 1` both sides, we get

`(1 + 1)^(15) = text()^( 15)C_0 + text()^( 15)C_1 + ... + text()^( 15)C_(15)`

`=> 2^(15) = text()^( 15)C_0 + text()^( 15)C_1 + ..... + text()^( 15)C_(15)`

Hence, Statement `2` is correct
Correct Answer is `=>` (C) Both 1 and 2
Q 2329245111

What is ` sum_(r = 0)^n C(n,r)` equal to?
NDA Paper 1 2013
(A)

`2^n - 1`

(B)

`n`

(C)

`n!`

(D)

`2^n`

Solution:

Given, ` sum_(r = 0)^n C (n, r)`

`= C (n, 0) + C(n, 1) + C(n, 2) + ... + C(n, n)`

`= text()^nC_0 + text()^nC_1 + text()^nC_2 + ... + text()^nC_n = (1+ 1)^n = 2^n`
Correct Answer is `=>` (D) `2^n`
Q 2521067821

Let `(1 + x)^(10) = sum_(r = 0)^(10) c_r x^r` and `(1 + x)^(7) = sum_(r = 0)^(7) d_r x^r` If ` P = sum_(r = 0)^(5) c_(2r)` and ` Q =
= sum_(r = 0)^(3) d_(2r+ 1)`, then `P/Q` is equal to
WBJEE 2012
(A)

`4`

(B)

`8`

(C)

`16`

(D)

`32`

Solution:

` P = sum_(r = 0)^(5) c_(2r)`

` = text()^(10)C_0 + text()^(10)C_2 + ... + text()^(10)C_(10) = 2^(10)/2 = 2^9`

` Q = sum_(r = 0)^(3) d_(2r+ 1) = d_1 + d_3 + d_5 + d_7`

` = text()^7C_1 + text()^7C_3 + text()^7C_5 + text()^7C_7 = 2^7/2 = 2^6`

`:. P/Q = 2^9/2^6 = 2^3 = 8`
Correct Answer is `=>` (B) `8`
Q 2468345205

If `1/(text()^(5)C_r) +1/(text ()^(6)C_r) =1/(text()^(4)C_r)`, then the value of `r` is
WBJEE 2016
(A)

`4`

(B)

`2`

(C)

`5`

(D)

`3`

Solution:

Given, `1/(text()^(5)C_r) +1/(text ()^(6)C_r) =1/(text()^(4)C_r)`

`=> (r! (5-r)!)/(5!) + (r! (6-r)!)/(6!) = (r!(4-r)!)/(4!)`

`=> ((5-r)!)/(5) + ((6-r)!)/30 = (4-r)!`

`=> (5-r)/5 +((6-r)(5-r))/30 =1`


`=> 30 - 6r + 30 - 11r + r^2 = 30`

`=> r^2 - 17r + 30 = 0`

`=> r^2 - 15r - 2r + 30 = 0`

`=> r(r -15) -2(r- 15) = 0`

`=> (r -2)(r -15) = 0`

`:. r=15,2`

Hence, the value of `r` is `2`.
Correct Answer is `=>` (B) `2`
Q 2580034817

The number of irrational terms in the
binomial expansion of `(3^(1/5) + 7^(1/3))^100` is
WBJEE 2015
(A)

`90`

(B)

`88`

(C)

`93`

(D)

`95`

Solution:

General term of `(3^(1/5) + 7^(1/3))^100` is given by

`T_(r+1) = text()^(100)C_r (3^(1/5))^(100-r) (7^(1/3) )^r`

`= text()^(100)C_r * 3^((100-r)/5) * 7^(r/3)`

For a rational term, `(100 -r)/5 ` and `r/3` must be integer.

Hence, `r=0, 15, 30, 45, 60, 75, 90`

`:.` There are seven rational terms.

Hence, number of irrational terms

`= 101-7=94`
Q 2523101041

`text()^(15)C_3 + text()^(15)C_5 +......+ text()^(15)C_15` will be equal to
WBJEE 2011
(A)

`2^14`

(B)

`2^(14) -15`

(C)

`2^(14) + 15`

(D)

`2^(14) -1`

Solution:

We know,

` text()^(15)C_1 + text()^(15)C_3 + text()^(15)C_5 +......+ text()^(15)C_15 = 2^(15-1)`

`:. text()^(15)C_3 + text()^(15)C_5 +......+ text()^(15)C_15 = 2^(14) -15`
Correct Answer is `=>` (B) `2^(14) -15`
Q 2408634508

The sum of the series

`text()^20 C_0 +text()^20 C_1+text()^20 C_2-text()^20 C_3+.........+ text()^20 C_10` is


BCECE Stage 1 2016
(A)

`-text()^20 C_10`

(B)

`1/2 text()^20 C_10`

(C)

`0`

(D)

`text()^20 C_10`

Solution:

On putting `x =-1` in

`(1+x)^20 =text()^20 C_0+text()^20 C_1x+......text()^20 C_10 x^10 +.....+ text()^20 C_20 x^20`

we get,

`0= text()^20 C_0-text()^20 C_1+.......-text()^20 C_9+text()^20 C_10- text()^20 C_11+.......+text()^20 C_20`

`=> 0=text()^20 C_0-text()^20 C_1+.........-text()^20 C_9+ text()^20 C_10`

`=> text()^20 C_10=2(text()^20 C_0-text()^20 C_1+...........+ text()^20 C_10)`

`=> text()^20 C_0- text()^20 C_1+............+ text()^20 C_10=1/2 text()^20 C_10`
Correct Answer is `=>` (B) `1/2 text()^20 C_10`
Q 2532191932

If `text()^(n)C_4 , text()^(n)C_5` and `text()^(n)C_6` are in AP, then `n` is
WBJEE 2011
(A)

`7` or `14`

(B)

`7`

(C)

`14`

(D)

`14` or `21`

Solution:

Since, ` text()^(n)C_4 , text()^(n)C_5` and `text()^(n)C_6` are in AP

`:. 2 text()^(n)C_5= text()^(n)C_4 + text()^(n)C_6`

`=> 2 xx (n!)/(5! (n-5)!)= (n!)/(4! (n-4)!) + (n!)/((n-6)! 6!)`

`=> 2/(5 (n-5)) =1/((n-4)(n-5)) +1/30`

`=> 2/(5 (n-5)) =1/((n-4)(n-5)) +1/30`

`=> 2/(5(n-5)) = (30+ n^2 -9n +20)/(30 (n-4)(n-5))`

`=>12 (n-4) = n^2 -9n +50`

`=n^2 -21 n +98 = 0`

`=> (n-14)(n-7)=0`

`= n =7, 14`
Correct Answer is `=>` (A) `7` or `14`
Q 2531567422

The sum of the series ` 1 + 1/2 text()^nC_1 + 1/3 text()^nC_1+ ... + 1/(n + 1) text()^nC_n` is equal to
WBJEE 2012
(A)

` (2^(n+1) - 1)/(n + 1)`

(B)

`(3(2^n - 1))/(2n)`

(C)

` (2^n + 1)/( n + 1) `

(D)

`(2^n + 1)/( 2n)`

Solution:

` 1 + 1/2 text()^nC_1 + 1/3 text()^nC_1+ ... + 1/(n + 1) text()^nC_n`

` = 1/(n + 1) [ (n + 1) + ((n + 1)n)/(2!) + ((n + 1)n( n -1))/(3!) + .. + 1]`

` = 1/(n + 1) [ text()^(n+1)C_1 + text()^(n+1)C_2 + ... + text()^(n+1)C_(n+1) ]`

` = 1/(n + 1) [ text()^(n+1)C_0 + text()^(n+1)C_1 + text()^(n+1)C_2 + ... + text()^(n+1)C_(n +1) - 1]`

` [ text()^(n+1)C_0 = 1 ]`

` = 1/(n + 1) (2^(n+1) - 1)`
Correct Answer is `=>` (A) ` (2^(n+1) - 1)/(n + 1)`

Application of binomial theorem

Q 2349345213

Let `n` be a positive integer and
`(1+x)^n = a_0 + a_1x + a_2x^2 + ... + a_nx^n`,
then what is `a_0 + a_1 + a_2 + ... + a_n` equal to?
NDA Paper 1 2013
(A)

`1`

(B)

`2^n`

(C)

`2^(n -1)`

(D)

`2^(n + 1)`

Solution:

Given that, `(1 + x)^n = a_0 + a_1x + a_2 x^2 + ... + a_n x^n`

Put `x = 1`

`(1 + 1)^n = a_0 + a_1 + a_2 + ... + a_n`

`:. a_0 + a_1 + a_2 + ... + a_n = 2^n`
Correct Answer is `=>` (B) `2^n`
Q 2369445315

What is the sum of the coefficients in the
expansion of `(1 + x)^n`?
NDA Paper 1 2013
(A)

`2^n`

(B)

`2^n -1`

(C)

`2^n + 1`

(D)

`n + 1`

Solution:

For the sum of the coefficients in the expansion of

`(1+x)^n`.

Put `x = 1 => (1 + x)^n = (1+1)^n = 2^n`

which is the required sum of the coefficients.
Correct Answer is `=>` (A) `2^n`
Q 2349645513

In the expansion of `(1 + x)^n`, what is the sum of
even Binomial coefficients?
NDA Paper 1 2012
(A)

`2^n`

(B)

`2^(n-1)`

(C)

`2^(n+1)`

(D)

None of these

Solution:

We know that,

`(1 + x)^n = text()^nC_0 + text()^nC_1x + text()^nC_2 x^2 + text()^nC_3 x^3 + text()^nC_4x^4 + ... + text()^nC_n x^n`

`(1- x)^n = text()^nC_0 - text()^nC_1x+ text()^nC_2x^2 - text()^nC_3 x^3 + text()^nC_4x^4 - ... +(-1)^n C_n x^n`

`(1 + x)^n + (1- x)^n = 2· text()^nC_0 + 2· text()^nC_2 x^2 + 2· text()^nC_4 x^4 + ...`

Put `x = 1`,

`=> (1+1)^n + (1-1)^n = 2{ text()^nC_0 + text()^nC_2 + text()^nC_4 + ... }`

`=> text()^nC_0 + text()^nC_2 + text()^nC_4 + .. = 1/2 (2^n + 0)`

` :. text()^nC_0 + text()^nC_2 + text()^nC_4 + .. = 2 ^(n -1)`
Correct Answer is `=>` (B) `2^(n-1)`
Q 2369256115

For all `n in N, 2^(4n) - 15n - 1` is divisible by
NDA Paper 1 2011
(A)

`125`

(B)

`225`

(C)

`450`

(D)

None of the above

Solution:

Given, `2^(4n) -15n - 1`

`(2^4)^n = (16)^n = (15 + 1)^n = (1 + 15)^n`

`= text()^nC_0 (15)^0 (1)^n + text()^nC_1 (1)^(n- 1) (15)^1 + text()^nC_2 (1)^(n - 2) (15)^2 + ...`

`= 1 + n·15 + text()^nC_2 (15)^2 + text()^nC_3 (15)^3 + ...`

` (2^4)^n - (15n) -1 =(15)^2 ( text()^nC_2 + text()^nC_3 ·15 + ... )`

` (2^4)^n - (15n) -1 = 225 . d , AA d in N`

Hence, `(2^4)^n - (15n) - 1` is divisible by `225`.
Correct Answer is `=>` (B) `225`
Q 2416712679

Larger of `99^50 + 100^50` and `101^ 50` is
UPSEE 2010
(A)

`101^50`

(B)

`99^50 +100^50`

(C)

both are equal

(D)

None of these

Solution:

We have

`100^50 -99^50=(100+1)^50-(100-1)^50`

`=100^50 (1+ 1/100)^50 -100^50 (1- 1/100)^50`

`=100^50 [2{text()^50 C_1 * 1/100+text()^50 C_3 (1/100)^2+.....}]`

`=(101^50 -99^50)/(100^50)`

`1+2{text()^50 C_3 (1/100)^2+........}`

`=> (101^50-99^50)/(100^50) > 1 +` (positive quantity) ` > 1`

`=> 101^50 -99^50 > 100^50`

`=> 101^50 > 100^50 +99^50`
Correct Answer is `=>` (A) `101^50`
Q 2405191968

`(100)^(50) + (99)^(50)`
UPSEE 2014
(A)

`< (101)^(50)`

(B)

`< (101)`

(C)

`> (101)^(50)`

(D)

`> (101)`

Solution:

Here, `(101)^(50) = (100 + 1)^(50) = 100^(50) + text()^(50)C_1 100^(49)`

`+ text()^(50)C_2 100^(48) + .... + 1` ... (i)

and `(99)^(50) = (100 - 1)^(50) = 100^(50) - text()^(50)C_1 100^(49)`

`+ text()^(50)C_2 100^(48) - ... + 1` ... (ii)

On subtracting Eq. (ii) from Eq. (i), we get

`(101)^(50) - (99)^(50) = 2 { text()^(50)C_1 100^(49) + text()^(50)C_3 100^(47) + ... }`

`=2 xx text()^(50)C_1 100^(49) + (2 xx text()^(50)C_3 xx 100^(47) + ... )`

`= 100 xx 100^(49) +` A positive number `> 100^(50)`

`=> (101)^(50) - (99)^(50) > (100)^(50)`

`=> (101)^(50) > (100)^(50) + (99)^(50)`

or `(100)^(50) + (99)^(50) < (101)^(50)`
Correct Answer is `=>` (A) `< (101)^(50)`
Q 2431812722

The last two digits of the number `3^400` are
UPSEE 2015
(A)

`81`

(B)

`43`

(C)

`29`

(D)

`01`

Solution:

Consider, `3^400 = (3^4)^100`

`=(81) ^100=(1+80)^100`

`=text()^100 C_0+text()^100 C_1 80+text()^100 C_2(80)^2
+......+ text()^100C_100(80)^100`

So, last two digits are `01`.
Correct Answer is `=>` (D) `01`
Q 2512591439

`(2^(3n)-1)` will be divisible by `(∀ n ϵ N)`
WBJEE 2010
(A)

`25`

(B)

`8`

(C)

`7`

(D)

`3`

Solution:

`2^(3n) = (8)^n = (1+7)^n`


` = text()^(1+)C_0+text()^nC_1 7+text()^nC_2 7^2 + .............+ text()^nC_n 7^n`



`=> 2^(3n)-1 = 7[text()^nC_1+text()^nC_2 7+........+text()^nC_n 7^(n-1)]`


`therefore` divisible by `7`
Correct Answer is `=>` (C) `7`
 
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