Solution:

Coefficient of `x^17` in `(x - 1) (x - 2) ... (x - 18)`

`= - (1 + 2 + 3 + ... +18)`

`= - (171)= -171`

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Solution:

The original set contains `2n + 1` element. The subsets containing more than `n` elements are
sets containing `n + 1, n + 2, n + 3, ... , (2n + 1)` elements.


`:.` Required number of subsets

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

`= text()^(2n+1)C_(n) + text()^(2n+1)C_(n-1) +.......+ text()^(2n+1)C_(0)`

`= text()^(2n+1)C_(0) + text()^(2n+1)C_(1 ) +..........+ text()^(2n+1)C_(n)`

`=1/2 [ (1+1)^(2n+1)] =1/2 [2^(2n +1)] = 2^(2n)`

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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`

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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`

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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`

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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`

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Solution:

We know that

`(1+x)^n=text()^n C_0+text()^n C_1x+text()^n C_2x^2+........+ text()^n C_nx^n`...................(i)

and `(x+1)^n= text()^n C_0x^n+ text()^n C_0x^(n-1) + text()^n C_2x^(n-2)+.......+ text()^n C_n`...........(ii)

On multiplying Eqs. (i) and (ii), we get

`(1+x)^(2n)=( text()^n C_0 + text()^n C_1 x+ text()^n C_2x^2+.......+ text()^n C_n x^n)xx (text()^n C_0 x^n + text()^n C_1x^(n-1)+ text()^n C_2x^(n-2) +.....+ text()^n C_n)`

Coefficient of `x ^n` in RHS

`={ text()^n C_0)^2+(text()^n C_1)^2+......+ (text()^n C_n)^2`

and coefficient of `x ^n` in LHS =` text()^(2n) C_n`

`:. (text()^n C_0 )^2 + (text()^n C_1)^2+ ....+ (text()^n C_n)^2= (2n!)/(n!n!)`

`=> (text()^n C_1)^2+....+ (text()^n C_n)^2= ((2n)!)/(n!n!)-1`

`=text()^(2n) C_n-1`

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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`

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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`

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Solution:

`∵ (1 + x^2)^(40) · (x^2 + 2 + 1/x^2)^(-5)`

`= (1 + x^2)^(20) . xx^(10)`

`= (1+ x^2)^(30). xx^(20)`

`:.` Coefficient of `x^(10)` in `(1 + x^2)^(30) · x^(10)`

`=>` Coefficient of `x^(10)` in `(1 + x^2)^(30)` is `text()^(30)C_5` or

` text()^(30)C_(25)`

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Solution:

` text()^(n+1)C_2 + 2 ( text()^(2)C_2 + text()^(3)C_2 + text()^(4)C_2 + .........+ text()^(n)C_2 )`

`= text()^(n+1)C_2 + 2 ( text()^(3)C_3 + text()^(3)C_2 + text()^(4)C_2 + ...... + text()^(n)C_2 )`

`= text()^(n+1)C_2 + 2 ( text()^(4)C_3 + text()^(4)C_2 + .......... + text()^(n)C_2)`

`= text()^(n+1)C_2 + 2 ( text()^(5)C_3 + text()^(5)C_2 + ........)`

`= text()^(n+1)C_2 + 2 text()^(n+1)C_3`

`= text()^(n+2)C_3 + text()^(n+1)C_3`

`= ((n+2)(n+1)n)/6 + ( (n+1)(n)(n-1))/6`

`= (n (n+1)(2n +1))/6`

`= 1^2 + 2^2 + 3^2 + ... + n^2`

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Solution:

`text()^(n-1)C_3+text()^(n-1)C_4 >text()^nC_4`


`=> text()^nC_4 > text()^nC_3` `[because text()^nC_r+text()^nC_(n+1) = text()^(n+1)C_(r+1)]`


`=> (n!)/(4!(n-4)!) > (4!)/(3!(n-3)!)`


`=> 1/4 > 1/(n-3)`


`=> n-3 > 4 => n > 7`

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Solution:

`(1 + x)^n = C_0 + C_1x + C_2x^2 + ... + C_n x^n`

`=> n(1 + x)^(n - 1) = C_1 + 2 * C_2x + .....+ n * C_n x^n` ... (1)


`=> n * 2^(n - 1) = C_1 + 2 * C_2 + 3 * C_3 + ...+ n * C_n`

`= aC_0 + (a + d)C_1+ (a + 2d)C_2 + ...(n + 1)` terms

`= a(C_0 + C_1 + C_2 + ...+ C_n) + d(C_1 + 2 * C_2+ 3 * C_3 + ... + n * C_n)`

`= a * 2^n + d * n 2^(n - 1)`

`= 2^n/(n+1) * (n+1)/2 (2a +nd)`

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

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Solution:

We have,

Coefficient of `x^n` in `(1 + x) (1 - x)^n`

`=` Coefficient of `x^n` in `(1 -x)^n`

`+` Coefficient of `x^(n - 1)` in `(1 -x)^n`

` = (-1)^n text()^nC_n + (-1)^(n -1) text()^nC_(n -1) = (-1)^n (1 - n)`

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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!)`

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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`.

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Solution:

The given series is in GP. Hence, its sum

` S = (1{(1+x)^(20 +1) -1})/((1 + x )-1) = ( (1+x)^(21) - 1)/x`

Therefore, the required coefficient of `x^(10)` in

the expansion of `((1 + x)^(21) - 1)/x`

`=` Coefficient of `x^(11)` in the expansion of

`(1+x)^(21) - 1`

` = text()^(21)C_(11)`

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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`.

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Solution:

Let the number of terms, `n =2m`

Now, by condition


`(text(Largest coefficient in) (1 + x)^n)/(text(Second largest coefficient in) (1 + x)^n) = 11/10`



` => (text()^(2m)C_m)/(text()^(2m)C_(m-1)) = 11/10`


`=> ((m-1)!(m+1)!)/(m!m!) = 11/10`


`=> 1/m (m(m-1)!(m+1)!)/(m!m!) = 11/10`


`=> (m+1)/m (m!m!)/(m!m!) = 11/10`

`=> 10m+10 = 11m `

`=> m = 10`

`therefore n = 20`
Hence, total number of term
`= n + 1 = 20 + 1 = 21`

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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`

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