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`

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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`

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Solution:

No of formed triangle `= 445`

`text()^15 C_3 - text()^n C_3 = 445`

`(15 * 14 * 13)/(3 * 2 *1) -(n(n-1) (n-2))/(3 * 2 * 1) = 445`

`- text()^n C_3 = 445-455`

`text()^n C_3 =10 = text()^5 C_3`

`n=5`

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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`

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Solution:

Let A be the set of natural numbers `(<= 1000)` which are divisible by 10,

B be the set of natural numbers `(<= 1000)` which are divisible by 15 and

C be the set of natural numbers `(<= 1000)` which are divisible by 25.

Then, `n(A) = [1000/10] = 100 ; n(B) = [1000/15]= 66` and

`n(c) = [1000/25] =40`, where `[*]` denotes the greater integer function.

Now, `n(A nn B)= n` (set of numbers which are divisible by both 10 and 15)

= n (set of numbers divisible by 30) `= [1000/30] =33`

Similarly, `n(A nn C) = n` (set of numbers divisible by 50)

`=[ 1000/75] =20`

and `n(A nn B nn C) = n` (set of numbers divisible by 10, 15 and 25)

= n (set of numbers divisible by 150)

`= [ 1000/150 ] =6`

`:. n(A uu B uu C)= n(A) + n(B) + n(C)- n(A nn B) - n(B nn C)- n(C nn A)+ n(A nn B nn C)`

`= 100 + 66 + 40 - 33 - 13 - 20 + 6 = 146`

Hence, required numbers `= n(U)- n(A uu B uu C)`
`= 1000 -146 = 854`

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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`

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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`

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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`

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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`

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Solution:

We have, `(x -1) (x- 2) (x- 3) ... (x -100)`

`= x^(100) - (1 + 2 + 3 + ..... + 100) x^(99) + ......`

:. Coefficient of `x^(99) = - (100)/2 [2 + 99] = -50 xx 101`

` = - 5050`

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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)`

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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.

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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`

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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`

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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`

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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.

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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

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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)`

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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`

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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`

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Solution:

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

`:.` Total term `= 20 + 1 = 21`

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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.

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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)`

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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`

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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)`

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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`.

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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`

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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`

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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`

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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`

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Solution:

Let `x = 8^(17) => x = 2^(51)`

Taking log on both sides of above equation, we get

`log x = 51 log 2`

`= 51 xx 0.3010 = 15.381`

`:.` Number of terms in `8^(17) = 15 + 1 = 16`.

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