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1.

Find a if 17th and 18th terms in the expansion of`(2+a)^(50)`are equal.A. `1//3`B. `1//2`C. `1`D. None of these

Answer» Correct Answer - C
N/a
Discussion
2.

Find the value of `1/(81^n)-(10)/(81^n)^(2n)C_1+(10^2)/(81^n)^(2n)C_2-(10^3)/(81^n)^(2n)C_3++(10^(2n))/(81^n)`.

Answer» Correct Answer - `1`
We have
`(1)/(81^(n)) - (10)/(81^(n)).^(2n)C_(1)+(10^(2))/(81^(n)).^(2n)C_(2)- (10^(3))/(81^(n)) .^(2n)C_(3)+"……."+(10^(2n))/(81^(n))`
`= 1/(81^(n))[.^(2n)C_(0) - .^(2n)C_(1)10^(1) + .^(2n)C_(2)10^(2)-.^(2n)C_(3)10^(3)+"……"+.^(2n)C_(2n)10^(2n)]`
`= (1)/(81^(n))[1-10]^(2n)`
`= ((-9)^(2n))/(81^(n))= (81^(n))/(81^(n)) = 1`
Discussion
3.

Prove that `^m C_1^n C_m-^m C_2^(2n)C_m+^m C_3^(3n)C_m-=(-1)^(m-1)n^mdot`

Answer» `.^(m)C_(1).^(n)C_(m)-.^(m)C_(2).^(2n)C_(3)+.^(m)C_(3).^(3n)C_(m)-"...."+(-1)^(m-1).^(m)C_(m).^(mn)C_(m)`
`= "Coefficient of" x^(m) " in"`
`.^(m)C_(1)(1+x)^(n)-.^(m)C_(2)(1+x)^(2n)+.^(m)C_(3)(1+x)^(3n)-"...."+(-1)^(m-1).^(m)C_(m)(1+x)^(mn)`
`=` Coefficient of `x^(m)` in
`.^(m)C_(0) - [.^(m)C_(0) - .^(m)C_(1)(1+x)^(n)+.^(m)C_(2)(1+x)^(2n)-"...."+(-1)^(m).^(m)C_(m)(1+x)^(mn)]`
`=` Coefficient of `x^(m)` in `[1-{1-(1+x)^(n)}^(m)]`
`=` Coefficient of `x^(m)` in `[1-{-nx-.^(n)C_(2)x^(2)-.^(n)C_(3)x^(3)-"......"-.^(n)C_(n)x^(n)}^(m)]`
`= - (-n)^(m)`
`= (-1)^(m-1)n^(m)`
Discussion
4.

Find a, b and n in the expansion of `(a+b)^n`if the first three terms of the expansion are 729, 7290 and 30375, respectively.

Answer» The first three term of the expansion are given as `729`, `7290`, and `30375`, respectively.
Therefore, we get
`T_(1) = .^(n)C_(0)a^(n-0)b^(0) = a^(n) = 729 " "(1)`
`T_(2) = .^(n)C_(1)a^(n-1)b^(1) = na^(n-1) b = 7290 " " (2)`
`T_(3) = .^(n)C_(2)a^(n-2)b^(2) = (n(n-1))/(2) a^(n-2) b^(2) = 30375" " (3)`
Dividing (2) by (1) , we get
`(na^(n-1)b)/(a^(n)) = (7290)/(729)`
or `(nb)/(a) = 10 " " (4)`
Dividing (3) by (2), we get
`(n(n-1)a^(n-2)b^(2))/( 2na^(n-1)b) = 30375/(7990)`
or `((n-1)b)/(a) = 25/3 " " (5)`
Dividing (4) by (5), we get
`(n)/(n-1) = 6/5`
or `n= 6`
Substituting `n = 6` in (1), we get
`a^(6) = 729` or `a = 3`
From (4), we get
`(6b)/(3) = 10` osr `b = 5`
Thus, `a = 3, b = 5` and `n = 6`.
Discussion
5.

Thesum of coefficients of integral powers of x in the binomial expansion of `(1-2sqrt(x))^(50)`is:(1) `1/2(3^(50)+1)`(2) `1/2(3^(50))`(3) `1/2(3^(50)-1)`(4) `1/2(2^(50)+1)`A. `1/2(3^(50) + 1)`B. `1/2(3^(50))`C. `1/2(3^(50) - 1)`D. `1/2(2^(10) + 1)`

Answer» Correct Answer - A
`(1-2sqrt(x))^(50) = C_(0)-C_(1)2sqrt(x)+C_(2)(2sqrt(x))^(2)-"...."+C_(50)+(2sqrt(x))^(50)`
`(1+2sqrt(x))^(50) = C_(0) + C_(1)(2sqrt(x))+C_(2)(2sqrt(x))^(2) + "......" + C_(50)(2sqrt(x))^(50)`
Putting `x = 1`, we get
`:. (3^(50)+1)/(2)= C_(0) + C_(2)(2)^(2) + "...."`
Discussion
6.

Thesum of coefficients of integral powers of x in the binomial expansion of `(1-2sqrt(x))^(50)`is:(1) `1/2(3^(50)+1)`(2) `1/2(3^(50))`(3) `1/2(3^(50)-1)`(4) `1/2(2^(50)+1)`A. `(1)/(2) (3^(50) + 1)`B. `(1)/(2) (3^(50))`C. `(1)/(2) (3^(50) - 1)`D. `(1)/(2) (3^(50) + 1)`

Answer» Let `T_(t + 1)` be the general term in the expansion of `(1 - 2 sqrt(x))^(50)`
`:. T_(r + 1) = .^(50)C_(r ) (1)^(50 - r) (-2x^(1//2))^(r ) = .^(50)C_(r ) 2^(r ) x^(r//2) (-1)^(r )`
For the integral power of x,r should be even integer.
`:.` Sum of coefficientss `= underset(r = 0)overset(25)(sum) .^(50)C_(2r) (2)^(2r)`
`= (1)/(2) [(1 + 2)^(50) + (1 - 2)^(50)] = (1)/(2) (30^(50) + 1)`
Alternate solution
we have
`(1 - 2 sqrt(2))^(50) = C_(0) - C_(1) 2 sqrt(x) + C_(2) (sqrt(2x))^(2 + ... + C_(50) (2 sqrt(x))^(50)` .... (i)
`(1 + 2 sqrt(x))^(50) = C_(0) + C_(1) 2 sqrt(x) + C_(2) (2 sqrt(x))^(2) + ... + C_(50) (2 sqrt(x))^(50)` ..... (iii)
On adding Eqs. (i) and (ii) we get
`(1 - 2sqrt(x))^(50) + (1 + 2 sqrt(x))^(50)`
`= 2 [C_(0) + C_(2) (2 sqrt(x))^(2) + ... + C_(50) (2 sqrt(x))^(50)]`
`implies ((1 - 2 sqrt(x))^(50) + (1 + 2 sqrt(1))^(50))/(2)`
`= C_(0) + C_(2) (2 sqrt(x))^(2) + .... + C_(50) (2 sqrt(x))^(50)`
On putting x - 1, we get
`((1 - 2 sqrt(1))^(50) + (1 + 2 sqrt(1))^(50))/(2) = C_(0) + C_(2) + ... + C_(50) (2)^(50)`
`implies ((-1)^(50) + (3)^(50))/(2) = C_(0) + C_(2) (2)^(2) + ... + C_(50) (2)^(50)`
`implies (1 + 3^(50))/(2) = C_(0) + C_(2) (2)^(2) +... + C_(50) (2)^(50)` .
Discussion
7.

Thesum of coefficients of integral powers of x in the binomial expansion of `(1-2sqrt(x))^(50)`is:(1) `1/2(3^(50)+1)`(2) `1/2(3^(50))`(3) `1/2(3^(50)-1)`(4) `1/2(2^(50)+1)`

Answer» `(1- 2 sqrtx)^50 = .^50c_0 - .^50c_1 (2 sqrtx)^1 + .^50c_2(2 sqrtx)^2- .^50c_3(2 sqrtx)^3 + .^50c_4 (2sqrtx)4 .....`
`s= .^50c_0 + .^50c_2 *2^2 + .^50 c_4 (2)^4 + .... + .^50c_502^50`
`(1+x)^50 = 1+ .^50C_1x^1 + .^50c_2x^2+...`
`x=2,-2`
`(1+2)^50 = 1 +.^50c_1 (2) + .^50c_2(2)^2 + ..`
`(1-2)^50 = 1- .^50c_1(2) + .^50c_2(2)^2 - .^50C_3(2)^3+ .....`
now,`3^50 + 1 = 2[.^50C_0 + .^50c_2 (2)^2 + .^50c_4(2)^4+ ..]`
`(3^50 +1)/2 = .^50C_0 + .^50c_2(2) + .^50c_4 (2)^4+...`
so, option 4 is correct
answer
Discussion
8.

In a binomial distribution `B(n , p=1/4)`, if the probability of at least one success isgreater than or equal to `9/(10)`, then n isgreater than(1) `1/((log)_(10)^4-(log)_(10)^3)`(2) `1/((log)_(10)^4+(log)_(10)^3)`(3) `9/((log)_(10)^4-(log)_(10)^3)`(4) `4/((log)_(10)^4-(log)_(10)^3)`

Answer» `0,1,2,3,4,..... , n`
`P= 1- `all fail
`S(P) = 1/4`
`S(F) = 1 - S(P)`
`= 1-1/4`
`=3/4`
`P= 1- (3/4)^n >= 9/10`
`-9/10 >= (3/4)^n`
`1/10 >= (3/4)^n`
`-1 >= n log_10 3/4`
`n <= -1/(log_10 3/4)`
`<= -1/(log_10 3 - log_10 4) `
`n >= 1/(log_10 4 - log_10 3) `
Answer
Discussion
9.

Least positive integer just greater than `(1+0. 00002)^(50000)`is.

Answer» Correct Answer - 3
`(1+0.0002)^(50000) = (1+1/(50000))^(50000)`
Now we know that `2 le (1+1/n)^(n) lt 3 AA n ge 1`
`rArr ` Least integer is 3
Discussion
10.

Find the sum of the roots (real or complex) of the equation `x^2001 + (1/2-x)^2001=0` .A. `2000`B. `2001`C. `1000`D. `500`

Answer» Correct Answer - D
`(d)` `x^(2001)-["^(2001)C_(0)x^(2001)-^(2001)C_(1)x^(2000)*(1)/(2)+^(2001)C_(2)x^(1999)*(1)/(4)...]=0`
`:. ^(2001)C_(1)x^(2000)*(1)/(2)-^(2001)C_(2)x^(1999)*(1)/(4)...=0`
sum of the roots `=^(2001)C_(2)*(1)/(4)*(2)/("^(2000)C_(1))=(2001*2000)/(4*2001)=500`
Discussion
11.

If the three consecutive in the expansion of `(1+x)^n`are 28, 56, and 70, then the value of `n`is.

Answer» Correct Answer - 8
Let the three cosecutive coefficient be `.^(n)C_(r-1)=28,.^(n)C_(r)=56` and `.^(n)C_(r+1)=70,`
so that `(.^(n)C_(r))/(.^(n)C_(r-1))=(n-r+1)/ ( r)=(56)/(28)=2` and `(.^(n)C_(r+1))/(.^(n)C_(r))=(n-r)/(r+1)=(70)/(56)=(5)/(4)`
This gives n+1 =3r and 4n-5=9r. Therefore, `(4n-5)/(n+1)=3 or n=8`
Discussion
12.

If `A_(i,j)` be the coefficient of `a^i b^j c^(2010-i-j)` in the expansion of `(a+b+c)^2010`, thenA. `A_(i,i)` is defined for `i ge 1010`B. `A_(i,j)=A_(j,i)`C. `A_(2i,3i)` is defined for `i ge 405`D. `A_(0,1)=2000`

Answer» Correct Answer - B
`(b)` Clearly `a_(i,j)=(2010!)/(i!j!(2010-i-j)!)`
and `a_(j,i)=(2010!)/(j!i!(2010-i-j)!)`
Hence , `a_(i,j)=a_(j,i)`
Discussion
13.

The coefficient of `x^70` in the product `(x-1)(x^2-2)(x^3-3)....(x^12-12)` isA. `4`B. `6`C. `8`D. `12`

Answer» Correct Answer - A
`(a)` The highest power of `x=1+2+3+….+12=78` to get coefficient of `x^(70)`, we have to omit the factors containing `x^(8)`
`(1)` Product of the constant terms of `(x-1)(x^(7)-7)=7`
`(2)` Product of the constant terms of `(x^(2)-2)(x^(6)-6)=12`
`(3)` Product of the constant terms of `(x^(3)-3)(x^(5)-5)=15`
`(4)` Product of the constant terms of `(x-1)(x^(2)-2)(x^(5)-5)=-10`
`(5)` Product of the constant terms of `(x-1)(x^(3)-3)(x^(4)-4)=-12`
Required coefficient `=7+12+15-10-12-8`
`=34-30=4`
Discussion
14.

The coefficient of `x^(301` ub the expansion of `(1+x)^(500)+x(1+x)^(499)+x^(2)(1+x)^(498)+…+x^(500)` isA. `"^(501)C_(301)`B. `"^(500)C_(301)`C. `"^(501)C_(300)`D. none of these

Answer» Correct Answer - A
`(a)` The given series
`S=(1+x)^(500)[1+(x)/(1+x)+((x)/(1+x))^(2)+....+((x)/(1+x))^(500)]`
`=(1+x)^(500)xx(1-((x)/(1+x))^(501))/(1-(x)/(1+x))`
`=(1+x)^(501)-x^(501)`
Hence, the coefficient of `x^(301)` in `S=^(501)C_(301)`.
Discussion
15.

The coefficient of `x^4`in `(x//2-3//x^2)`is`(405)/(256)`b. `(504)/(259)`c. `(450)/(263)`d. none of theseA. `(405)/(256)`B. `(504)/(259)`C. `(450)/(263)`D. None of these

Answer» The general term in `((x)/(2)-(3)/(x^(2)))` is
`t_(r+1)=(-1)^(r).^(10)C_(r)((x)/(2))^(10-r)((3)/(x^(2)))^(r)=(-r)^(r).^(10)C_(r).(3^(r))/(2^(10-r)).x^(10-3r)`
For coefficient of `x^(4)`, we put `10-3r=4`
`implies3r=6`
`impliesr=2`
`therefore`coefficient of `x^(4)` in `((x)/(2)-(3)/(x^(2)))^(10)=(-1)^(2.).^(10)C_(2).(3^(2))/(2^(8))`
`=(45xx9)/(256)=(405)/(256)`
Discussion
16.

Show that `9^(n+1)-8n-9`is divisible by 64, whenever n is a positive integer.

Answer» In order to showe that `9^(n+1) - 8n -9` is divisible by 64, it has to be proved that `9^(n+1) - 8n - 9 = 64 k`, where k is some natural number
`(1+8)^(n+1) = .^(n+1)C_(0) + .^(n+1)C_(1)(8)+.^(n+1)C_(2)(8)^(2)+"….."+.^(n+1)C_(n+1)(8)^(n+1)`
or `9^(n+1)= 1+(n+1)(8)+8^(2)[.^(n+1)C_(2)+.^(n+1)C_(3) xx 8 + "....."+.^(n+1)C_(n+1)(8)^(n-1)]`
or `9^(n+1)=9+8n+64[.^(n+1)C_(2)+.^(n+1)C+(3) xx 8 + "...."+.^(n+1)C_(n+1)(8)^(n-1)]`
or `9^(n+1) - 8n - 9 = 64 k`, where `k = .^(n+1)C_(2) + .^(n+1)C_(3) xx 8 + "......"+.^(n+1)C_(n+1)(8)^(n-1)` is a positive integers.
Discussion
17.

Find an approximation of `(0. 99)^5`using the first three terms of its expansion.

Answer» `0.99 = 1 - 0.01`
`:. (0.999)^(5) = (1-0.01)^(5)`
`= .^(5)C_(0)(1)^(5) - .^(5)C_(1) (1)^(4) - (-0.01) + .^(5)C_(2)(1)^(3)(0.01)^(2)` (Approximately)
` = 1-5(0.01) + 10(0.01)^(2)`
` = 1-0.05 + 0.001`
`= 1.001 - 0.05`
`= 0.951`
Thus, the value of `(0.99)^(5)` is approximately `0.951`.
Discussion
18.

Using binomial theorem, expand each of the following: `(3x^(2)-2ax+3a^(2))^(3)`

Answer» Correct Answer - `27x^(6)-54ax^(5)+ 117a^(2)x^(4)-116a^(3)x^(3) + 117a^(4)x^(2)-54a^(5)x+27a^(6)`
`(3x^(2) - 2ax + 3a^(2))^(3) = [(3x^(2) - 2ax)+3a^(2)]^(3)`
`=.^(3) C _(0) (3x^(2) - 2ax)^(3) + .^(3)C_(1) (3x^(2) - 2ax)^(2)(3a^(2))+.^(3)C_(2) (3x^(2)-2ax) (3a^(2))^(2) + .^(3)C_(3)(3a^(2))^(3)`
`=[.^(3)C_(0)(3x^(2))^(3) -.^(3)C_(1)(3x^(2))^(2) xx2ax+.^(3)C_(2)(3x^(2)) xx (2ax)^(2) - .^(3)C_(3)(2ax)^(3)]`
`+(9x^(4) + 4a^(2)x^(2) - 12ax^(3)) (3a^(2)) + (27a^(4)) (3x^(2) - 2ax) + 27a^(6)`
`=27x^(6) - 54ax^(5) + 36a^(2)x^(4) - 8a^(3)x^(3) + 81a^(2)x^(4) +36a^(4)x^(2) - 108a^(3) x^(3) + 81a^(4)x^(2)-54a^(5) x + 27a^(6)`
`=27x^(6) - 54ax^(5) +117a^(2)x^(2) -116a^(3)x^(3) + 117a^(4)x^(2) - 54a^(5)x = 27a^(6).`
Discussion
19.

Find the cofficient of the term independent of x in the expansion of `((x+1)/(x^(2/3)-x^(1/3)+1)-(x-1)/(x-x^1/2))^10`

Answer» Correct Answer - 210
`(x+1)/((x^(2//3) - x^(1//3))+1) - (x-1)/(x-x^(1//2))`
`= ((x^(1//3) +)(x^(2//3) - x^(1//3)+1))/(x^(2//3) - x^(1//3) + 1) - (x^(1//2) + 1)/(x^(1//2))`
`= x^(1//3) + 1 - 1 - x^(-1//2)`
` = x^(1//3) - x^(-1//2)`
`:. ((x+1)/(x^(2//3) - x^(1//3) + 1) - (x-1)/(x-x^(1//2)))^(10) = (x^(1//3) - x^(-1//2))^(10)`
Let `T_(r+1) = .^(100)C_(r) (x^(1//3))^(10-r) (-1)^(r) (-x^(-1//2))^(r )`
For the term independent of x,
`(10-r)/(3) - (r )/(2) = 0`
`rArr 20 - 2r - 3r = 0`
`rArr r = 4`
So, required coefficient `= .^(10)C_(4)(-1)^(4) = 210`
Discussion
20.

If `|x|

Answer» Correct Answer - 1
Since `1+2z+3x^(2)+4x^(3)+"……"oo= (1+x)^(2)`, we have
`(1+2x+3x^(2)+4x^(3)+"……"oo)^(1//2) = [(1-x)^(-2)]^(1//2)`
`= (1-x)^(-1)`
`= 1+x+x^(2)+"…."+x^(n) + "…."oo`
Therefore, the coefficient of `s^(n)` is `1`.
Discussion
21.

If p is a real number and the middle term in the expansion of `((p)/(2) + 2)^(8)` is 1120, then find the value of p.

Answer» Given expansion is `((p)/(2) + 2)^(8)`
Hence, `n = 8`
Since, this Binomial expansion has only one middle term i.e., `((8)/(2) + 1)th = 5 th` term
`T_(5) = T_(4 + 1) = .^(8)C_(4) ((p)/(2))^(8 - 4) .2^(4)`
`rArr 1120 = .^(8)C_(4) p^(4) .2^(-4) 2^(4)`
`rArr 1120 = (8 xx 7 xx 6 xx 5 xx 4!)/(4! xx 4 xx 3 xx 2 xx 1)p^(4)`
`rArr 1120 = 7 xx 2 xx 5 xx p^(4)`
`rArr p^(4) = (1120)/(70) = 16 rArr p^(4) = 2^(4)`
`rArr p^(2) = 4 rArr p = -+ 2`
Discussion
22.

If the coefficients of 2nd, 3rd and 4th terms in the expansion of`(1+x)^n` are in A.P., then find the value of n.A. 2B. 7C. 11D. 14

Answer» The expansion of `(1 + x)^(n) " is " .^(n)C_(0) + .^(n)C_(1) x + .^(n)C_(2) x^(2) + .^(n)C_(3) x^(3) + .... + .^(n)C_(n) x^(n)`
`:.` Coefficient of 2nd term `= .^(n)C_(1)`,
Coefficient of 3 rd term `= .^(n)C_(2)`,
and coefficient of 4 th term `= .^(n)C_(3)`
Given that, `.^(n)C_(1), .^(n)C_(2) " and " .^(n)C_(3)` are in AP.
`:. 2 .^(n)C_(2) = .^(n)C_(1) + .^(n)C_(3)`
`rArr 2[((n)!)/((n - 2)! 2!)] = ((n)!)/((n - 1)!) + ((n)!)/(3! (n - 3)!)`
`rArr (2.n (n - 1) (n - 2)!)/((n - 2)! 2!) = (n (n - 1)!)/((n - 1)!) + (n (n - 1) (n - 2) (n - 3)!)/(3.2.1 (n - 3)!)`
`rArr n (n - 1) = n + (n (n - 1) (n - 2))/(6)`
`rArr 6n - 6 = 6 + n^(2) - 3n + 2`
`rArr n^(2) - 9n + 14 = 0`
`rArr n^(2) - 7n - 2n + 14 = 0`
`rArr n(n - 7) - 2 (n - 7) = 0`
`:. n = 2 " or " n = 7`
Sine, `n = 2` is not possible
`:. n = 7`
Discussion
23.

Find the coefficient of `x^4`in the expansion of `(1+x+x^2+x^3)^(11)dot`

Answer» Given, expansion `= (1 + x + x^(2) + x^(3))^(11) = [(1 + x) + x^(2) (1 + x)]^(11)`
`= [(1 + x) (1 + x^(2))]^(11) = (1 + x)^(11) (1 + x^(2))^(11)`
Now, above expansion becomes
`= (.^(11)C_(0) + .^(11)C_(1) x + .^(11)C_(2) x^(2) + .(11)C_(3) x^(3) + .^(11)C_(4) x^(4) +....) (.^(11)C_(0) + .^(11)C_(1) x^(2) + .^(11)C_(2) x^(4) + ...)`
`= (1 + 11 x + 55 x^(2) + 165 x^(3) + 330 x^(4) +...) (1 + 11 x^(2) + 55 x^(4) +...)`
`:.` Coefficient of `x^(4) = 55 + 605 + 330 = 990`
Discussion
24.

Prove that the coefficient of `x^n`in the expansion of `1/((1-x)(1-2x)(1-3x))i s1/2(3^(n+2)-2^(n+3)+1)dot`

Answer» We have `(1)/((1-x)(1-2x)(1-3x))`
` = (1)/(2(1-x))- (4)/(1-2x)+(9)/(2(1-3x))` [By resolving into partial fractions]
`= 1/2(1-x)^(-1)-4(1-2x)^(-1)+9/2(1-3x)^(-1)`
`= 1/2(1+x+x^(2)+"….."+x^(n)+"…..")-4(1+2x+(2x)^(2)+"…."+(2x)^(n)+"….")+9/2(1+(3x)+(3x)^(2)+"...."+(3x)^(n)+"......")`
`:.` Coefficient of `x^(n) = 1/2[1-8.2^(n)+9.3^(n)]`
`=1/2[1-2^(n+3)+3^(n+2)]`
Discussion
25.

If `(1 + x - 2 x^(2))^(6) = 1 + C_(1) x + C_(2) x^(2) + C_(3) x^(3) + …+ C_(12) x^(12)`, then the value of `C_(2) + C_(4) + C_(6) + …+ C_(12)` isA. 30B. 32C. 31D. none of these

Answer» Correct Answer - c
Discussion
26.

If the coefficient of 2nd, 3rd and 4thterms in the expansion of `(1+x)^(2n)`are in A.P. , show that `2n^2-9n+7=0.`A. 1.3B. 0.2C. 4D. -1

Answer» Correct Answer - b
Discussion
27.

If a. b, c and d are the coefficients of 2nd, 3rd, 4th and 5th terms respectively in the binomial expansion of `(1+x)^n`, then prove that `a/(a+b) + c/(c+d) = 2b/(b+c)`A. `(b)/(b+c)`B. `(b)/(2(b+c))`C. `(2b)/(b+c)`D. `(2c)/(b+c)`

Answer» Correct Answer - C
N/a
Discussion
28.

If the coefficients of 2nd, 3rd and 4th terms in the expansion of`(1+x)^n` are in A.P., then find the value of n.

Answer» Let the coefficients of 2nd, 3rd and 4th terms in the expansion of `(1+x)^n is .^C_1, .^nC_2, .^nC-3`.
According to given condition,
`2(.^nC_2)=.^nC_1+.^Nc_3`
Discussion
29.

Find the term in `(a/(sqrt(b))3+sqrt(b/(a3)))^(21)`which has the same power of `aa n dbdot`A. 9B. 10C. 8D. 6

Answer» Correct Answer - a
Discussion
30.

If the coefficient of 2nd, 3rd and 4thterms in the expansion of `(1+x)^(2n)`are in A.P. , show that `2n^2-9n+7=0.`

Answer» Given expansion is `(1 + x)^(2n)`
Now, coefficient of 2nd term `= .^(2n)C_(1)`
Coefficient of 3rd term `= .^(2n)C_(2)`
Coefficient of 4th term `= .^(2n)C_(3)`
Given that, `.^(2n)C_(1), .^(2n)C_(2) " and " .^(2n)C_(3)` are in AP.
Then, `2. .^(2n)C_(2) = .^(2n)C_(1) + .^(2n)C_(3)`
`rArr 2 [(2n (2n - 1) (2n - 2)!)/(2 xx 1 xx (2n - 2)!)] = (2n (2n -1)!)/((2n - 1)!) + (2n (2n -1) (2n - 2) (2n - 3)!)/(3! (2n - 3)!)`
`rArr n(2n - 1) = n + (n(2n - 1) (2n -2))/(6)`
`rArr n(12n - 6) = n (6 + 4n^(2) - 4n - 2n + 2)`
`rArr 12 n - 6 = (4n^(2) - 6 n + 8)`
`rArr 6(2n - 1) = 2 (2n^(2) - 3n + 4)`
`rArr 3(2n - 1) = 2n^(2) - 3n + 4`
`rArr 2n^(2) - 3n + 4 - 6n + 3 = 0`
`rArr 2n^(2) - 9n + 7 = 0`
Discussion
31.

Find the coefficient of `x^4`in the expansion of `(1+x+x^2+x^3)^(11)dot`A. 900B. 909C. 990D. 999

Answer» Correct Answer - c
Discussion
32.

If `9^7-7^9`is divisible b `2^n ,`then find the greatest value of `n ,w h e r en in Ndot`

Answer» We have
`9^(7)-7^(9) = (1+8)^(7) - (1-8)^(9)`
`= (1+.^(7)C_(1)8^(1)+.^(7)C_(2)8^(2)+"….."+.^(7)C_(7)8^(7)) - (1-.^(9)C_(1)8^(1)+.^(9)C_(2)8^(2)-"……."-.^(9)C_(9)8^(9))`
`= 16xx8+64[(.^(7)C_(2)+"….."+.^(7)C_(7)8^(5))-(.^(9)C_(2)-"……"-.^(9)C_(9)8^(7))]`
`= 64k` (where k is some integer )
Therefore, `9^(7) - 7^(9)` is divisible by `64`.
Discussion
33.

If the coefficients of three consecutive terms in the expansion of `(1+x)^n`are in the ratio 1:7:42, then find the value of `ndot`

Answer» Let `(r+1)` th, `(r+2)`t h, and `(r+3)`th be the three consecutive terms. Then,
`.^(n)C_(r): .^(n)C_(r+1)=1:7:42`
Now, `(.^(n)C_(r))/(.^(n)C_(r+1))=1/7` or `(r+1)/(n-r)=1/7` or `n-8r = 7" "(1)`
and `(.^(n)C_(r+1))/(.^(n)C_(r+2))=7/42` or `(r+2)/(n-r-1)=1/6`
`rArr n-7r=13" "(2)`
Solving (1) and (2), we get `n = 55`.
Discussion
34.

17. If the coefficients of 2nd, 3rd and 4th terms in the expansion of ` (1+x)^(2n)` are in A.P.. Show that `2n^2-9n+7=0`A. `2n^(2) + 9n+7 = 0`B. `2n^(2) - 9n + 7 = 0`C. `2n^(2) - 9n - 7 = 0`D. none of these

Answer» Correct Answer - b
Discussion
35.

Find the sum of coefficients in the expansion of `(x-2y+3z)^n`is 128, then find the greatest coefficients in the expansion of `(1+x)^ndot`

Answer» Correct Answer - `.^(7)C_(3)` or `.^(7)C_(4)`
Sum of the coefficient in the expansion of
`(x-2y+3z)^(n)` is `(1-2+3)^(2) = 2^(n)` (putting `x = y = z = 1`)
`:. 2^(n) = 128` or `n = 7`
Therefore, the greatest coefficient in the expansion of `(1+x)^(7)` is `.^(7)C_(3)` or `.^(7)C_(4)` because both are equal to `35`.
Discussion
36.

The coefficient of the (r-1)th, rth and (r+1)th terms in the expansion of `(x + 1)^n` are in the ratio 1:3:5. Find both n and r

Answer» We have
`(x+1) (n) = (1+x)^(n) `
`=.^(n)C_(0) + .^(n)C _(1) x + .^(n) C _ (2) x^(2) + ...+ .^(n) C _(r-2) x ^(r-2) + .^(n) C _(r-1) x^(r-1)`
`+.^(n) C _(r) x ^(r)....`
`:. T _((r-1)) = T _((r-2) + 1) = .^(n)C(r-2) x ^(r-2)," "` ...(i)
`T_(r) = T _((r-1) + 1) = .^(n) C _(r-1) x ^(r-1)," "` ...(ii)
`T_(r+1)= .^(n)C_(r) x ^(r)." "`...(iii)
It is given that
` " coeff. of " T _(r-1) : "coeff. of " T_(r) : "coeff . of " T_(r+1)=1: 3 : 5`
`rArr .^(n)C _((r-2)) : .^(n)C_((r-1)) : .^(n) C _(r) = 1: 3 : 5 `
`rArr (.^(n) C _(r-2) )/(.^(n)C _(r-1)) = 1/3 and (.^(n) C _(r-1))/(.^(n) C r ) = 3/5`.
But,`(.^(n)C_(r-2))/(.^(n)C_(r-1))=1/3 rArr (n!)/(((r-2)!*)(n-r+2)!)xx ((r-1)!*(n-r+1)!)/(n!) =1/3`
` rArr ((r-1))/((n-r+2))=1/3`
`rArr n-4r = -5 " "` ...(i)
And, `(.^(n) C _(r-1))/(.(n) C _(r)) = 3/5 rArr r/ ((n -r+1)) = 3/5 `
`rArr 3n - 8r = - 3." "` ...(ii)
Solving (i) and (ii) , we get `n= 7 and r=3.`
Hence, `n=7and r=3.
Discussion
37.

If A and B are the coefficients of `x^n` in the expansion `(1 + x)^(2n)` and `(1 + x)^(2n-1)` respectively, thenA. `A=B`B. `2A=B`C. `A=2B`D. None of these

Answer» Correct Answer - C
N/a
Discussion
38.

If the coefficients of `x^7 and x^8` in the expansion of `[2 +x/3]^n` are equal, then the value of n is : (A) 15 (B) 45 (C) 55 (D) 56A. 15B. 45C. 55D. 60

Answer» Correct Answer - C
N/a
Discussion
39.

If the coefficients of the `(n+1)^(t h)`term and the `(n+3)^(t h)`term in the expansion of `(1+x)^(20)`are equal , then the value of `n`is`10`b. `8`c. `9`d. none of theseA. PB. `P+1`C. `P+2`D. `P+3`

Answer» Correct Answer - B
N/a
Discussion
40.

If the sum of the coefficients in the expansion of `(a+b)^n`is 4096, then the greatest coefficient in the expansion is`924`b. `792`c. `1594`d. none of theseA. `792`B. `924`C. `1048`D. `2096`

Answer» Correct Answer - B
N/a
Discussion
41.

If the coefficients of `x^(2)` and `x^(3)`are both zero, in the expansion of the expression `(1 + ax + bx^(2)) (1 - 3x)^15` in powers of x, then the ordered pair (a,b) is equal to(A)` (28, 315)`(B)`(-21, 714) `(C) `(28, 861)`(D) `(-54, 315)`A. (28, 315)B. (-21, 714)C. (28, 861)D. (-54, 315)

Answer» Given expression is `(1 + ax + bx^(2)) (1 - 3x)^(15)`. In the expanison of binominal `(1 - 3x)^(15)`. The `(r + 1)` th term is
`T_(r + 1) = .^(15)C_(r ) - (3x)^(r ) = .^(15)C_(r ) (-3)^(r ) x^(r )`
Now, coefficient of `x^(2)` in the expansion of `(1 + ax + bx^(2)) (1 - 3x)^(15)` is
`.^(15)C_(2) (-3)^(2) + a .^(15)C_(1) (-3)^(1) + b .^(15)C_(0) (-3)^(0) = 0` (given)
`implies (105 xx 9) - 45a + b = 0`
`implies 45 a - b = 945`
Similarly, the coefficient of `x^(3)`, in the expanison of
`(1 + ax + bx^(2)) (1 - 3x)^(15)` is
`.^(15)C_(3) (-3)^(3) + a .^(15)C_(2) (-3)^(2) + b .^(15)C_(1) (-3)^(1) = 0`
`implies - 12285 + 945 a - 45 b = 0`
`implies 63 a - 3b = 819`
`implies 21a - b = 273`
From Eqs. (i) and (ii) We get
`24 a = 672 implies a = 28`
So, b = 315
`implies (a, b) = (28, 315)`
Discussion
42.

The second, third and fourth terms in the binomial expansion `(x+a)^n`are 240, 720 and 1080, respectively. Find x, a and n.A. 15B. 20C. 10D. 55

Answer» Correct Answer - d
Discussion
43.

The second, third and fourth terms in the binomial expansion `(x+a)^n`are 240, 720 and 1080, respectively. Find x, a and n.

Answer» By the binomial expansion , we have
`(x+a)^(n) = .^(n) C _(0) x ^(n)+ .^(n) C _ (1) x^(n-1) a+ .^(n)C_(2) x ^(n-2) a^(2) + . ^(n) C _(3) x ^(n-3) a^(3) + ...`
`rArr (x+a)^(n) = x^(n) + nx ^(n-1) a + 1/2 n ( n-1) x^((n-2)) a^(2)`
`+ 1/6 n ( n-1) ( n-2) x^(n-3 a^(3) + ...`
` :. T_(2) = 240 rArr nx^(n-1) a = 240, " "` ...(i)
`T_(3) = 720 rArr n(n-1) x^(n-2) a^(2) = 1440," "` ...(ii)
`T_(4) = 1080 rArr n ( n-1) (n-2) x^(n-3) a^(3) = 6480. " "` ...(iii)
On dividing (ii) by (i) , we get
`((n-1)a)/x = 1440/240 rArr a/x = 6/((n-1)). " "` ...(iv)
On dividing (iii) by (ii), we get
`((n-2)a)/x = 6480/1440 rArr a/ x = 9/(2(n-2))." "` ...(v)
Equating the values of `a/x` from (iv) and (v), we get
`6/(n-1) = 9 /2(n-2) rArr 12(n-2) = 9 ( n-1) rArr 3n = 15 rArr n=5.`
Putting n=5 in (i) we get `x^(4) a = 48" "` ...(vi)
Putting n=5 (iv) , we get `a/x = 6/4 rArr a/x = 3/2 rArr a = 3/2 x.`
Putting `a=3/2 x ` in (vi), we get
`x^(4) 3/2 x = 48 rArr x^(5) = 48 xx 2/3 = 32 = 2^(5) rArr x= 2 .`
Hence, ` x=2, a=3, and n=5.`
Discussion
44.

If the coefficient of `x^7` in `(ax^2+1/(bx))^11` is equal to the coefficient of `x^7` in `(ax-1/(bx^2))^11` thenA. `a+b=1`B. `a-b=1`C. `ab=1`D. `a/b=1`

Answer» Correct Answer - C
For `(ax^(2)+(1/(bx)))^(11)`.
`T_(r+1)=.^(13)C_(r)(ax^(2))^(11-r)(1/(bx))^(r) = .^(11)C_(r ) a^(11-r) (1)/(b^(r)) x^(22-3r)`
For `x^(7)`,
`22-3r = 7`
or `3r = 15`
or `r = 5`
`rArr T_(6) = .^(11)C_(5)a^(6)(1)/(b^(5)) x^(7)`
`rArr` Coefficient of `x^(7)` is `.^(11)C_(5) (a^(6))/(b^(5))`
Similarly, coefficient of `x^(-7)`in `(ax- (1/(bx^(2))))^(11)` is `.^(11)C_(6)(a^(5))/(b^(6))`.
Given that `.^(11)C_(5) (a^(6))/(b^(5)) = .^(11)C_(6) (a^(5))/(b_(6))`
`rArr a= 1/b`
or `ab = 1`
Discussion
45.

The value of cube root of 1001 upto five decimal places isA. 10.03333B. 10.00333C. 10.00033D. none of these

Answer» `(1001)^(1//3)=(1000+1)^(1//3)=10(1+1/(1000))^(1//3)=10{1+1/(3)1/(1000)+(1//3(1//3-1))/(2!)(1)/(1000^2)+....}`
=10{1+0.00003333-0.00000011+…..}=10.00333
Discussion
46.

If `((1-3x)^(1//2)+(1-x)^(5//3))/(sqrt(4-x))` is approximately equal to `a+bx` for small values of `x`, then `(a,b)=`A. `(1,(35)/(24))`B. `(1,-(35)/(24))`C. `(2,(35)/(12))`D. `(2,-(35)/(12))`

Answer» Correct Answer - B
`(b)` `((1-3x)^(1//2)+(1-x)^(5//3))/(2[1-(x)/(4)]^(1//2))`
`=([1+(1)/(2)(-3x)]+[1+(5)/(3)(-x)])/(2[1+(1)/(2)(-(x)/(4))])`
(Neglecting higher powers of `x`)
`=([1-(19)/(12)x])/([1-(x)/(8)])`
`=[1-(19)/(12)x][1-(x)/(8)]^(-1)`
`=[1-(19)/(12)x][1+(x)/(8)]=1-(35)/(24)x`
then `a+bx=1-(35)/(24)ximpliesa=1`, `b=-(35)/(24)`
Discussion
47.

In the expansion of `(3^(-x//4)+3^(5x//4))^n`the sum of binomial coefficient is 64 and term with the greatestbinomial coefficient exceeds the thirdby `(n-1)`, the value of `x`must be`0`b. `1`c. `2`d. `3`

Answer» Correct Answer - A
To get sum of coefficients put `x = 0` Given that sum of coefficient is
`2^(n) = 64`
or `n = 6`
The greatest binomial coefficeint is `.^(6)C_(3)`.
No given that
`T_(4) - T_(3) = 6-1 = 5`
`rArr .^(6)C_(3) (3^(-x//4))^(3)(3^(5x//4))^(3)-.^(6)C_(2)(3^(-x//4))^(4)(3^(5x//4))^(2) = 5`.
Which is satisfied by `x = 0`
Discussion
48.

Assuming `x`to be so small that `x^2`and higher power of `x`can be neglected, prove that

Answer» We have,
`((1+3/4x)^(-4)(16-3x)^(1//2))/((8+x)^(2//3))=((1+3/4x)^(-4)(16)^(1//2)(1-(3x)/(16))^(1//2))/(8^(2//3)(1+x/8)^(2//3))`
`= (1+3/4x)^(-4) (1-(3x)/(16))^(1//2) (1+(x)/(8))^(-2//3)`
`{1+(-4) (3/4x)}{1+1/2 ((-3x)/(16))}{1+(-2/3)(x/8)}`
`= (1-3x)(1-3/32x)(1-x/12)`
`= (1-3x-3/(32)x)(1-x/12)` [neglecting `x^(2)`]
`= (1-99/32x)(1-x/12)=1-(99)/(32)x-(x)/(12)` [neglecting `x^(2)`]
`1 - (305)/(96) x`
Discussion
49.

Find the numerically Greatest Term In the expansion of `(3-5x)^15` when x=1/5

Answer» We have `(3-5x)^(15)`
For `|(T_(r+1))/(T_(r))|=(15-r+1)/(r) |-(5x)/(3)|ge 1`
`rArr (16-r)/(r).(5xx1/5)/(3)ge1 ""("Putting" x = 1//5)`
`rArr 16 - r ge 3r`
`rArr le 4`
Hence, `T_(4)` and `T_(5)` numerically the greatest terms,
`T_(4) = .^(15)C_(3)3^(15-3)(-5x)^(3)`
`= (-455 xx 3^(12) xx 5^(3))x^(3)`
and `T_(5) = .^(15)C_(4)3^(15-4)(-5x)^(4)`
`= (455 xx 3^(12) xx 5^(3))x^(4)`
Also , for numerical value,
`|T_(4)| = |T_(5)| = 455 xx 3^(12) 5^(3)`
Discussion
50.

For which of the following values of `x ,5t h`term is the numerically greatest term in the expansion of `(1+x//3)^(10)`,a. -2 b. 1.8 c. `2`d. `-1. 9`A. -2B. 1.8C. 2D. -19

Answer» Correct Answer - A::B::C::D
Let `T_(5)` is numerically the greatest term in the expansion of`(1+x//3)^(10)`
Then, `|(T_(5))/(T_(4))| ge 1` and `|(T_(6))/(T_(5))| le 1`
Now, `(T_(r+1))/(T_(r)) = (10-r+1)/(r ) x/3`
`rArr |(7)/(4) xx x/3| ge 1` and `|6/5 xx x/3| le 1`
`rArr |x| gt 12/7` and `|x| le 5/2 " "(1)`
`rArr 12/7 le |x| le 5/2`
`rArr x in [-5/7 , - 12/7] uu [12/7,5/2]`
Discussion