JEE Mathematics Binomial Theorem MCQs Set F

Question. If 'a' be the sum of the odd terms & 'b' the sum of the even terms in the expansion of \( (1+x)^n \), then \( (1-x^2)^n \) equals
(a) \( a^2 - b^2 \)
(b) \( a^2 + b^2 \)
(c) \( b^2 - a^2 \)
(d) None of the options
Answer: (c) \( b^2 - a^2 \)
Solution:
\( (1 + x)^n = \sum_{r=0}^{n} {}^nC_r x^r = ({}^nC_0 x^0 + {}^nC_2 x^2 + \dots) + ({}^nC_1 x^1 + {}^nC_3 x^3 + \dots) \)
\( \begin{cases} T_1 + T_3 + T_5 + \dots = a \\ T_2 + T_4 + T_6 + \dots = b \end{cases} \)
\( \therefore (1 - x)^n = ({}^nC_0 x^0 + {}^nC_2 x^2 + \dots) - ({}^nC_1 x^1 + {}^nC_3 x^3 + \dots) = a - b \)
\( (1 - x^2)^n = (1 + x)^n (1 - x)^n = (a + b) (a - b) = a^2 - b^2 \)

Question. Given that the term of the expansion \( (x^{1/3} - x^{-1/2})^{15} \) which does not contain \( x \) is \( 5m \) where \( m \in \mathbb{N} \), then \( m \) equals
(a) 1100
(b) 1010
(c) 1001
(d) None of the options
Answer: (c) 1001
Solution:
\( (x^{1/3} - x^{-1/2})^{15} \)
\( T_{r+1} = {}^{15}C_r (x^{1/3})^{15-r} (-x^{-1/2})^r \)
\( T_{r+1} = {}^{15}C_r (-1)^r (x)^{\frac{15-r}{3}} \cdot x^{-\frac{r}{2}} = {}^{15}C_r (-1)^r (x)^{\frac{30-5r}{6}} \)
Given if \( \frac{30 - 5r}{6} = 0 \) then \( T_{r+1} = 5m, m \in \mathbb{N} \)
\( \Rightarrow r = 6 \quad \Rightarrow \quad T_7 = 5m \)
\( T_7 = {}^{15}C_6 (-1)^6 = \frac{15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2} = 5 \cdot 7 \cdot 13 \cdot 11 = 5 \cdot (1001) \Rightarrow m = 1001 \)

Question. In the binomial \( (2^{1/3} + 3^{-1/3})^n \), if the ratio of the seventh term from the beginning of the expansion to the seventh term from its end is \( 1/6 \), then \( n \) equals
(a) 6
(b) 9
(c) 12
(d) 15
Answer: (b) 9
Solution:
\( 7^{\text{th}} \) term of \( (2^{1/3} + 3^{-1/3})^n \) is \( T_7 = {}^nC_6 2^{\frac{n-6}{3}} 3^{-2} \)
From begining
\( 7^{\text{th}} \) term of \( (2^{1/3} + 3^{-1/3})^n \) from the end
\( = 7^{\text{th}} \) term of \( (3^{-1/3} + 2^{1/3})^n \) from the begining
\( = T'_7 = {}^nC_6 3^{-\frac{(n-6)}{3}} 2^2 \)
given that \( \frac{T_7}{T'_7} = \frac{1}{6} \)
\( \Rightarrow \frac{{}^nC_6 2^{\frac{n-6}{3}} 3^{-2}}{{}^nC_6 3^{-\frac{(n-6)}{3}} 2^2} = \frac{1}{6} \Rightarrow \frac{2^{\frac{n-6}{3} - 2}}{3^{2 - \frac{(n-6)}{3}}} = \frac{1}{6} \)
\( \Rightarrow \frac{1}{(2 \cdot 3)^{2 - (\frac{n-6}{3})}} = \frac{1}{6} \Rightarrow 2 - \frac{n-6}{3} = 1 \)
\( \Rightarrow 3 = n - 6 \Rightarrow n = 9 \)

Question. Let \( n \) be a positive integer. Then of the following, the greatest term is
(a) \( \left(1 + \frac{1}{4n}\right)^{4n} \)
(b) \( \left(1 + \frac{1}{3n}\right)^{3n} \)
(c) \( \left(1 + \frac{1}{2n}\right)^{2n} \)
(d) \( \left(1 + \frac{1}{n}\right)^n \)
Answer: (a) \( \left(1 + \frac{1}{4n}\right)^{4n} \)
Solution:
\( \left(1 + \frac{1}{4n}\right)^{4n} = {}^{4n}C_0 \left(\frac{1}{4n}\right)^0 + {}^{4n}C_1 \left(\frac{1}{4n}\right)^1 + {}^{4n}C_2 \left(\frac{1}{4n}\right)^2 + \dots \)
\( = 1 + 1 + \frac{1\left(1 - \frac{1}{4n}\right)}{2!} + \dots + \frac{1\left(1 - \frac{1}{4n}\right)\left(1 - \frac{2}{4n}\right)\dots\left(1 - \frac{4n-1}{4n}\right)}{4n!} \)
\( < 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{(4n)!} \)
\( \left(1 + \frac{1}{3n}\right)^{3n} = 1 + 1 + \frac{1\left(1 - \frac{1}{3n}\right)}{2!} + \dots + \frac{1\left(1 - \frac{1}{3n}\right)\left(1 - \frac{2}{3n}\right)\dots\left(1 - \frac{3n-1}{3n}\right)}{(3n)!} \)
\( < 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{(3n)!} \)
\( \left(1 + \frac{1}{2n}\right)^{2n} = 1 + 1 + \frac{1\left(1 - \frac{1}{2n}\right)}{2!} + \dots + \frac{1\left(1 - \frac{1}{2n}\right)\left(1 - \frac{2}{2n}\right)\dots\left(1 - \frac{2n-1}{2n}\right)}{(2n)!} \)
\( < 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{(2n)!} \)
\( \left(1 + \frac{1}{n}\right)^n = 1 + 1 + \frac{1\left(1 - \frac{1}{n}\right)}{2!} + \dots + \frac{1\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)\dots\left(1 - \frac{n-1}{n}\right)}{n!} \)
\( < 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{(n)!} \)
\( \left(1 + \frac{1}{n}\right)^n < \left(1 + \frac{1}{2n}\right)^{2n} < \left(1 + \frac{1}{3n}\right)^{3n} < \left(1 + \frac{1}{4n}\right)^{4n} \)

Question. If the coefficients of \( x^7 \) & \( x^8 \) in the expansion of \( \left[2 + \frac{x}{3}\right]^n \) are equal, then the value of \( n \) is
(a) 15
(b) 45
(c) 55
(d) 56
Answer: (c) 55
Solution:
\( T_{r+1} = {}^nC_r (2)^{n-r} \left(\frac{x}{3}\right)^r = {}^nC_r \frac{2^{n-r}}{3^r} x^r \)
Coeff \( T_8 = T_9 \Rightarrow {}^nC_7 \frac{2^{n-7}}{3^7} = {}^nC_8 \frac{2^{n-8}}{3^8} \)
\( \Rightarrow \frac{{}^nC_8}{{}^nC_7} = \frac{3 \cdot 2}{1} \Rightarrow \frac{n - 8 + 1}{8} = 6 \)
\( \Rightarrow n - 7 = 48 \Rightarrow n = 55 \)

Question. The expression \( \frac{1}{\sqrt{4x + 1}} \left[ \left( \frac{1 + \sqrt{4x + 1}}{2} \right)^7 - \left( \frac{1 - \sqrt{4x + 1}}{2} \right)^7 \right] \) is a polynomial in \( x \) of degree
(a) 7
(b) 5
(c) 4
(d) 3
Answer: (d) 3
Solution:
\( \frac{1}{\sqrt{4x + 1}} \left[ \left( \frac{1 + \sqrt{4x + 1}}{2} \right)^7 - \left( \frac{1 - \sqrt{4x + 1}}{2} \right)^7 \right] \)
\( = \frac{1}{2^7 \sqrt{4x + 1}} \left[ \sum_{r=0}^7 {}^7C_r (1)^{7-r} (\sqrt{4x + 1})^r - \sum_{r=0}^7 {}^7C_r (1)^{7-r} (-\sqrt{4x + 1})^r \right] \)
\( = \frac{1}{2^7 \sqrt{4x + 1}} [T_1 + T_2 + T_3 + \dots + T_8 - T_1 + T_2 - T_3 + \dots + T_8] \)
\( = \frac{1}{2^7 \sqrt{4x + 1}} [2(T_2 + T_4 + T_6 + T_8)] \)
\( = \frac{1}{2^6 \sqrt{4x + 1}} \left[ {}^7C_1 \sqrt{4x + 1} + {}^7C_3 (\sqrt{4x + 1})^3 + {}^7C_5 (\sqrt{4x + 1})^5 + {}^7C_7 (\sqrt{4x + 1})^7 \right] \)
\( = \frac{1}{2^6} \left[ {}^7C_1 + {}^7C_3 (4x + 1) + {}^7C_5 (4x + 1)^2 + {}^7C_7 (4x + 1)^3 \right] \)
= is the polynomial in x of degree 3

Question. Number of rational terms in the expansion of \( (\sqrt{2} + \sqrt[4]{3})^{100} \) is
(a) 25
(b) 26
(c) 27
(d) 28
Answer: (b) 26
Solution:
G.T. is \( T_{r+1} = {}^{100}C_r (2)^{\frac{100-r}{2}} (3)^{\frac{r}{4}} \)
The above term will be rational if exponent of 2 & 3 are integers.
i.e. \( \frac{100 - r}{2} \) and \( \frac{r}{4} \) must be integers
the possible set of r is = {0, 4, 8, 16, \dots, 100}
no. of rational terms is 26

Question. If \( n \in \mathbb{N} \) & \( n \) is even, then \( \frac{1}{1 \cdot (n - 1)!} + \frac{1}{3!(n - 3)!} + \frac{1}{5!(n - 5)!} + \dots + \frac{1}{(n - 1)!1!} \) equals
(a) \( 2^n \)
(b) \( \frac{2^{n-1}}{n!} \)
(c) \( 2^n n! \)
(d) None of the options
Answer: (b) \( \frac{2^{n-1}}{n!} \)
Solution:
If \( n \in \mathbb{N} \) & \( n \) is even then
\( \frac{1}{1 \cdot (n - 1)!} + \frac{1}{3!(n - 3)!} + \frac{1}{5!(n - 5)!} + \dots + \frac{1}{(n - 1)! 1!} \)
\( = \frac{1}{n!} [{}^nC_1 + {}^nC_3 + {}^nC_5 + \dots + {}^nC_{n-1}] \)
\( n \) is even \( \Rightarrow n - 1 \) is odd
\( {}^nC_{n-1} \) second Binomail coeff. from the end
\( = \frac{1}{n!} [{}^nC_1 + {}^nC_3 + {}^nC_5 + \dots + {}^nC_{n-1}] \)
\( = \frac{1}{n!} \cdot 2^{n-1} = \frac{2^{n-1}}{n!} \)

Question. The sum of the series \( (1^2 + 1) 1! + (2^2 + 1) \cdot 2! + (3^2 + 1) \cdot 3! + \dots + (n^2 + 1) \cdot n! \) is
(a) \( (n + 1) \cdot (n + 2)! \)
(b) \( n \cdot (n + 1)! \)
(c) \( (n + 1) \cdot (n + 1)! \)
(d) None of the options
Answer: (b) \( n \cdot (n + 1)! \)
Solution:
\( (1^2 + 1) 1! + (2^2 + 1) 2! + (3^2 + 1) 3! + \dots + (n^2 + 1) \cdot n! \)
\( T_n = \sum_{n=1}^{n} (n^2 + 1) n! \)
\( = \sum_{n=1}^{n} [n^2 + 3n + 2 - 3n - 1] n! \)
\( = \sum_{n=1}^{n} [(n + 2)(n + 1) - 3(n + 1) + 2] n! \)
\( = \sum_{n=1}^{n} [(n + 2)! - 3(n + 1)! + 2n!] \)
\( = 3! + 4! + 5! + \dots + n! + (n + 1)! + (n + 2)! \)
\( - 3[2! + 3! + 4! + 5! + \dots + n! + (n + 1)!] \)
\( + 2[1! + 2! + 3! + \dots + n!] \)
\( = [(n + 2)! + (n + 1)! - 3(n + 1)! - 3 \cdot 2! + 2 \cdot 1! + 2 \cdot 2!] \)
\( = (n + 2)! - 2(n + 1)! - 2! + 2 \cdot 1! \)
\( = (n + 1)! [n + 2 - 2] \)
\( = (n + 1)! \cdot n \)
\( = n \cdot (n + 1)! \)

Question. The last two digits of the number \( 3^{400} \) are
(a) 81
(b) 43
(c) 29
(d) 01
Answer: (d) 01
Solution:
\( 3^{400} = (3^2)^{200} = (9)^{200} = (10 - 1)^{200} \)
\( = {}^{200}C_0 10^{200} - {}^{200}C_1 10^{199} + {}^{200}C_2 10^{198} - \dots - {}^{200}C_{199} 10^1 + {}^{200}C_{200} 10^0 \)
\( = 100 \left[ {}^{200}C_0 10^{198} - {}^{200}C_1 10^{197} + {}^{200}C_2 10^{196} - \dots + {}^{200}C_{198} \right] - {}^{200}C_{199} 10^1 + {}^{200}C_{200} 10^0 \)
\( = (k)100 - 200 (10) + 1(1) \)
\( = (k - 20) 100 + 01 \)
so last two digits is 01

Question. The sum of the binomial coefficients of \( \left[2x + \frac{1}{x}\right]^n \) is equal to 256. The constant term in the expansion is
(a) 1120
(b) 2110
(c) 1210
(d) None of the options
Answer: (a) 1120
Solution:
\( C_0 + C_1 + C_2 + \dots + C_n = 2^n = 256 \)
\( \Rightarrow 2^n = 2^8 \Rightarrow n = 8 \)
\( T_{r+1} = {}^8C_r (2x)^{8-r} \left(\frac{1}{x}\right)^r = {}^8C_r 2^{8-r} x^{8-2r} \)
For Constant term \( \Rightarrow 8 - 2r = 0 \Rightarrow r = 4 \)
\( = {}^8C_4 2^4 = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2} 2^4 = 70 \times 16 = 1120 \)

Question. The sum of the co-efficients in the expansion of \( (1 - 2x + 5x^2)^n \) is 'a' and the sum of the co-efficients in the expansion of \( (1 + x)^{2n} \) is b. Then
(a) \( a = b \)
(b) \( a = b^2 \)
(c) \( a^2 = b \)
(d) \( ab = 1 \)
Answer: (a) \( a = b \)
Solution:
sum of coeff of \( (1 - 2x + 5x^2)^n = a \)
sum of coeff of \( (1 + x)^{2n} = b \)
put \( x = 1 \)
\( a = (1 - 2 + 5)^n = 4^n \) & \( b = (1 + 1)^{2n} = 2^{2n} = 4^n \)
\( a = b \)

Question. The sum of the co-efficients of all the even powers of \( x \) in the expansion of \( (2x^2 - 3x + 1)^{11} \) is
(a) \( 2 \cdot 6^{10} \)
(b) \( 3 \cdot 6^{10} \)
(c) \( 6^{11} \)
(d) None of the options
Answer: (b) \( 3 \cdot 6^{10} \)
Solution:
Let \( (2x^2 - 3x + 1) = a_0 + a_1x + a_2x^2 + a_3x^3 + \dots + a_{22}x^{22} \dots (1) \)
Put \( x = -1 \)
\( \Rightarrow a_0 - a_1 + a_2 - a_3 + \dots + a_{22} = (2 + 3 + 1)^{11} = 6^{11} \dots (2) \)
Put \( x = 1 \)
\( \Rightarrow a_0 + a_1 + a_2 + a_3 + \dots + a_{22} = 0 \dots (3) \)
adding (2) & (3)
\( \Rightarrow 2[a_0 + a_2 + a_4 + \dots + a_{22}] = 6^{11} \)
\( \Rightarrow a_0 + a_2 + a_4 + \dots + a_{22} = \frac{6^{11}}{2} = \frac{6 \cdot 6^{10}}{2} = 3 \cdot 6^{10} \)

Question. Set of values of \( r \) for which, \( {}^{18}C_{r - 2} + 2 \cdot {}^{18}C_{r - 1} + {}^{18}C_r \geq {}^{20}C_{13} \) contains
(a) 4 elements
(b) 5 elements
(c) 7 elements
(d) 10 elements
Answer: (c) 7 elements
Solution:
\( {}^{18}C_{r - 2} + 2 \cdot {}^{18}C_{r - 1} + {}^{18}C_r \geq {}^{20}C_{13} \)
\( \Rightarrow {}^{18}C_{r - 2} + {}^{18}C_{r - 1} + {}^{18}C_{r - 1} + {}^{18}C_r \geq {}^{20}C_{13} \)
\( \Rightarrow {}^{19}C_{r - 1} + {}^{19}C_r \geq {}^{20}C_{13} \Rightarrow {}^{20}C_r \geq {}^{20}C_{13} \)
\( \therefore {}^{20}C_{10} > {}^{20}C_{11} > {}^{20}C_{12} > {}^{20}C_{13} \)
& \( {}^{20}C_{10} > {}^{20}C_9 > {}^{20}C_8 > {}^{20}C_7 \)
\( r = 7, 8, 9, 10, 11, 12, 13 \Rightarrow \) Total 7 elements

Question. The greatest terms of the expansion \( (2x + 5y)^{13} \) when \( x = 10, y = 2 \) is
(a) \( {}^{13}C_5 \cdot 20^8 \cdot 10^5 \)
(b) \( {}^{13}C_6 \cdot 20^7 \cdot 10^4 \)
(c) \( {}^{13}C_4 \cdot 20^9 \cdot 10^4 \)
(d) None of the options
Answer: (c) \( {}^{13}C_4 \cdot 20^9 \cdot 10^4 \)
Solution:
\( (2x + 5y)^{13} \) greatest form for \( x = 10, y = 2 \)
\( \frac{n + 1}{\left|\frac{2x}{5y}\right| + 1} - 1 \leq r \leq \frac{n + 1}{\left|\frac{2x}{5y}\right| + 1} \)
\( \Rightarrow \frac{14}{\left|\frac{20}{10}\right| + 1} - 1 \leq r \leq \frac{14}{\left|\frac{20}{10}\right| + 1} \)
\( \Rightarrow \frac{14}{3} - 1 \leq r \leq \frac{14}{3} \)
\( \Rightarrow \frac{11}{3} \leq r \leq \frac{14}{3} \)
\( \Rightarrow 3.66 \dots \leq r \leq 4.666 \Rightarrow r = 4 \)
\( \Rightarrow T_5 = {}^{13}C_4 (20)^9 (10)^4 \)

Question. The binomial expansion of \( \left(x^k + \frac{1}{x^{2k}}\right)^{3n} \), \( n \in \mathbb{N} \) contains a term independent of \( x \)
(a) only if \( k \) is an integer
(b) only if \( k \) is a natural number
(c) only if \( k \) is rational
(d) for any real \( k \)
Answer: (d) for any real \( k \)
Solution:
\( \left(x^k + \frac{1}{x^{2k}}\right)^{3n} \), \( n \in \mathbb{N} \) Independent of \( x \)
\( T_{r+1} = {}^{3n}C_r (x^k)^{3n-r} \left(\frac{1}{x^{2k}}\right)^r \)
\( = {}^{3n}C_r x^{3nk - rk - 2kr} = {}^{3n}C_r x^{3k(n - r)} \)
For Constant term \( \Rightarrow 3k(n - r) = 0 \Rightarrow n = r \)
\( \therefore T_{r+1} = {}^{3n}C_n \) true for any real \( k \) or \( k \in \mathbb{R} \)

Question. \( \frac{C_0}{1} + \frac{C_1}{2} + \frac{C_2}{3} + \dots + \frac{C_{10}}{11} = \)
(a) \( \frac{2^{11}}{11} \)
(b) \( \frac{2^{11} - 1}{11} \)
(c) \( \frac{3^{11}}{11} \)
(d) \( \frac{3^{11} - 1}{11} \)
Answer: (b) \( \frac{2^{11} - 1}{11} \)
Solution:
\( \frac{C_0}{1} + \frac{C_1}{2} + \frac{C_2}{3} + \dots + \frac{C_{10}}{11} \)
\( = \sum_{r=0}^{10} \frac{{}^{10}C_r}{r + 1} = \sum_{r=0}^{10} \frac{1}{11} \frac{11}{r + 1} {}^{10}C_r = \frac{1}{11} \sum_{r=0}^{10} {}^{11}C_{r+1} \)
\( = \frac{1}{11} \left[ {}^{11}C_1 + {}^{11}C_2 + \dots + {}^{11}C_{11} \right] \)
\( = \frac{1}{11} \left[ {}^{11}C_0 + {}^{11}C_1 + {}^{11}C_2 + \dots + {}^{11}C_{11} - {}^{11}C_0 \right] \)
\( = \frac{1}{11} \left[ 2^{11} - 1 \right] = \frac{2^{11} - 1}{11} \)

Question. Let \( (5 + 2\sqrt{6})^n = p + f \) where \( n \in \mathbb{N} \) and \( p \in \mathbb{N} \) and \( 0 < f < 1 \) then the value, \( f^2 - f + pf - p \) is
(a) a natural number
(b) a negative integer
(c) a prime number
(d) an irrational number
Answer: (b) a negative integer
Solution:
\( P + f = (5 + 2\sqrt{6})^n \), \( P, n \in \mathbb{N} \), \( 0 < f < 1 \)
Let \( f' = (5 - 2\sqrt{6})^n = |\text{rational} - \text{irrational}|^n < 1 \)
\( P + f + f' = 2 [\text{Integer}] = \text{even integer} \)
\( \because f + f' = 1 \Rightarrow (f - 1) = -f' \)
Now \( f^2 - f + Pf - P = f(f - 1) + P(f - 1) \)
\( = (f - 1) (P + f) = -f' (P + f) \)
\( = -(5 - 2\sqrt{6})^n (5 + 2\sqrt{6})^n = - [5^2 - 2^2 \cdot 6]^n \)
\( = - [25 - 24]^n = -1 = \text{negative Interger} \)

Question. The coefficient of \( x^r(0 \leq r \leq n - 1) \) in the expression \( (x + 2)^{n-1} + (x + 2)^{n-2} \cdot (x + 1) + (x + 2)^{n-3} \cdot (x + 1)^2 + \dots + (x + 1)^{n-1} \) is
(a) \( {}^nC_r (2^r - 1) \)
(b) \( {}^nC_r (2^{n-r} - 1) \)
(c) \( {}^nC_r (2^r + 1) \)
(d) \( {}^nC_r (2^{n-r} + 1) \)
Answer: (b) \( {}^nC_r (2^{n-r} - 1) \)
Solution:
\( (x + 2)^{n - 1} + (x + 2)^{n - 2} (x + 1) + (x + 2)^{n - 3} (x + 1)^2 + \dots + (x + 1)^{n - 1} \)
\( = a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \dots + b^{n - 1} \)
\( = a^{n-1} \left[ 1 + \left(\frac{b}{a}\right) + \left(\frac{b}{a}\right)^2 + \dots + \left(\frac{b}{a}\right)^{n-1} \right] \)
\( = a^{n-1} \left[ \frac{1 \cdot \left(\left(\frac{b}{a}\right)^n - 1\right)}{\frac{b}{a} - 1} \right] = a^{n-1} \frac{a^n - b^n}{a-b} \cdot \frac{a}{a^n} \)
\( = \frac{(x + 2)^n - (x + 1)^n}{x + 2 - x - 1} = (2 + x)^n - (1 + x)^n \)
\( T_{r+1} \) term is \( {}^nC_r 2^{n - r} x^r - {}^nC_r x^r \)
coeff of \( x^r \) is \( {}^nC_r (2^{n - r} - 1) \)

Question. If \( (1 + x + x^2)^{25} = a_0 + a_1x + a_2x^2 + \dots + a_{50} \cdot x^{50} \) then \( a_0 + a_2 + a_4 + \dots + a_{50} \) is
(a) even
(b) odd & of the form 3n
(c) odd & of the form (3n – 1)
(d) odd & of the form (3n + 1)
Answer: (a) even
Solution:
\( (1 + x + x^2)^{25} = a_0 + a_1x + a_2x^2 + \dots a_{50}x^{50} \)
put \( x = 1 \)
\( a_0 + a_1 + a_2 + a_3 + \dots + a_{49} + a_{50} = 3^{25} \dots (i) \)
put \( x = -1 \)
\( a_0 - a_1 + a_2 - a_3 + \dots + a_{49} + a_{50} = (1 - 1 + 1)^{25} \dots (ii) \)
adding (i) & (ii)
\( 2[a_0 + a_2 + a_4 + \dots + a_{50}] = 3^{25} + 1 \)
\( a_0 + a_2 + a_4 + \dots + a_{50} = \frac{3^{25} + 1}{2} = \frac{(4 - 1)^{25} + 1}{2} \)
\( = \frac{{}^{25}C_0 4^{25} - {}^{25}C_1 4^{24} + {}^{25}C_2 4^{23} - \dots + {}^{25}C_{24} 4^1 - 1 + 1}{2} \)
\( = \frac{4 [^{25}C_0 4^{24} - {}^{25}C_1 4^{23} + \dots {}^{25}C_{24}]}{2} \)

Question. The co-efficient of \( x^4 \) in the expansion of \( (1 - x + 2x^2)^{12} \) is
(a) \( {}^{12}C_3 \)
(b) \( {}^{13}C_3 \)
(c) \( {}^{14}C_4 \)
(d) \( {}^{12}C_3 + 3 \cdot {}^{13}C_3 + {}^{14}C_4 \)
Answer: (d) \( {}^{12}C_3 + 3 \cdot {}^{13}C_3 + {}^{14}C_4 \)
Solution:
coef of \( x^4 \) in \( (1 - x + 2x^2)^{12} \)
\( = {}^{12}C_0 (1 - x)^{12} (2x^2)^0 + {}^{12}C_1 (1 - x)^{11} (2x^2)^1 + {}^{12}C_2 (1 - x)^{10} (2x^2)^2 + \text{above } x^4 \text{ powers} \)
terms of \( x^4 \)
\( = {}^{12}C_0 \cdot {}^{12}C_4 (-x)^4 + {}^{12}C_1 \cdot {}^{11}C_2 (-x)^2 2x^2 + {}^{12}C_2 \cdot {}^{10}C_0 4x^4 \)
\( = {}^{12}C_4 + 12 \cdot {}^{11}C_2 \cdot 2 + {}^{12}C_2 \cdot 4 \)
\( = {}^{12}C_4 + 2 \cdot 3 \cdot \frac{12}{3} \cdot {}^{11}C_2 + {}^{12}C_2 \cdot 4 \)
\( = {}^{12}C_3 + {}^{12}C_2 + 3({}^{12}C_2 + {}^{12}C_3) + {}^{12}C_3 + {}^{12}C_4 \)
\( = {}^{12}C_3 + 3 \cdot {}^{13}C_3 + {}^{13}C_3 + {}^{13}C_4 \)
\( = {}^{12}C_3 + 3 \cdot {}^{13}C_3 + {}^{14}C_4 \)

Question. If \( \sum_{k=1}^{n-r} {}^{n-k}C_r = {}^xC_y \) then
(a) \( x = n + 1 ; y = r \)
(b) \( x = n ; y = r + 1 \)
(c) \( x = n ; y = r \)
(d) \( x = n + 1 ; y = r + 1 \)
Answer: (b) \( x = n ; y = r + 1 \)
Solution:
\( \sum_{k=1}^{n-r} {}^{n-k}C_r = {}^xC_y \)
L.H.S. \( = {}^{n - 1}C_r + {}^{n - 2}C_r + {}^{n - 3}C_r + \dots + {}^rC_r \)
\( = {}^rC_r + {}^{r + 1}C_r + \dots + {}^{n - 2}C_r + {}^{n - 1}C_r \)
\( \left\{ {}^rC_r = {}^{r + 1}C_{r + 1} \right\} \)
\( = {}^{r + 1}C_{r + 1} + {}^{r + 1}C_r + {}^{r + 2}C_r + \dots + {}^{n - 1}C_r \)
\( = {}^{r + 2}C_{r + 1} + {}^{r + 2}C_r + \dots + {}^{n - 1}C_r \)
\( = {}^{r + 2}C_{r + 1} + \dots + {}^{n - 1}C_r \)
\( = {}^{n - 1}C_{r + 1} + {}^{n - 1}C_r \)
\( = {}^nC_{r + 1} = {}^xC_y \Rightarrow x = n, y = r + 1 \)

Question. Coefficient of \( \alpha^t \) in the expansion of \( (\alpha+p)^{m-1} + (\alpha+p)^{m-2} (\alpha+q) + (\alpha+p)^{m-3} (\alpha+q)^2 + \dots + (\alpha+q)^{m-1} \) where \( \alpha \neq -q \) and \( p \neq q \) is
(a) \( \frac{{}^mC_t (p^t - q^t)}{p - q} \)
(b) \( \frac{{}^mC_t (p^{m-t} - q^{m-t})}{p - q} \)
(c) \( \frac{{}^mC_t (p^t + q^t)}{p - q} \)
(d) \( \frac{{}^mC_t (p^{m-t} + q^{m-t})}{p - q} \)
Answer: (b) \( \frac{{}^mC_t (p^{m-t} - q^{m-t})}{p - q} \)
Solution:
Coeff of \( \alpha^t \) in
\( (\alpha + p)^{m - 1} + (\alpha + p)^{m - 2} (\alpha + q) + (\alpha + p)^{m - 3} (\alpha + q)^2 + \dots + (\alpha + q)^{m - 1} \)
\( \because a \neq -q, p \neq q \)
Let \( \alpha + p = x \) & \( \alpha + q = y \)
\( = x^{m - 1} + x^{m - 2} y + x^{m - 3} y^2 + \dots + y^{m - 1} \)
\( = x^{m - 1} \left[ 1 + \left(\frac{y}{x}\right) + \left(\frac{y}{x}\right)^2 + \dots + \left(\frac{y}{x}\right)^{m - 1} \right] \)
\( = x^{m-1} \left[ \frac{1 - \left(\frac{y}{x}\right)^m}{1 - \frac{y}{x}} \right] \)
\( = \frac{x^m - y^m}{x - y} \cdot x = \frac{(\alpha + p)^m - (\alpha + q)^m}{\alpha + p - \alpha - q} \)
\( = \frac{1}{(p - q)} \left[ (\alpha + p)^m - (\alpha + q)^m \right] \)
= coeff of \( \alpha^t = \left( \frac{{}^mC_t p^{m - t} - {}^mC_t q^{m - t}}{p - q} \right) \)

Question. The co-efficient of \( x^{401} \) in the expansion of \( (1 + x + x^2 + \dots + x^9)^{-1} \), \( (|x| < 1) \) is
(a) 1
(b) -1
(c) 2
(d) -2
Answer: (b) -1
Solution:
\( (1 + x + x^2 + \dots + x^9)^{-1} \quad (|x| < 1) \)
\( = \left[ \frac{1 \cdot (1 - x^{10})}{1 - x} \right]^{-1} = \frac{(1 - x)}{(1 - x^{10})} = (1 - x)(1 - x^{10})^{-1} \)
\( = (1 - x) [1 + (x^{10}) + (x^{10})^2 + (x^{10})^3 + \dots \infty] \)
\( = (1 - x) [1 + x^{10} + x^{20} + x^{30} + x^{40} + \dots + x^{400} + x^{410} + \dots \infty] \)
= coeff of \( x^{401} \) is \( (-1) \)

Question. Number of terms free from radical sign in the expansion of \( (1 + 3^{1/3} + 7^{1/7})^{10} \) is
(a) 4
(b) 5
(c) 6
(d) 8
Answer: (c) 6
Solution:

\( \therefore \) no. of terms are 6
Alter : \( {}^{10}C_r \left(1 + 3^{\frac{1}{3}}\right)^{10-0} \Rightarrow {}^{10}C_r \left(1 + 3^{\frac{1}{3}}\right)^r \)
should be r = 0, 3, 6, 9
\( {}^{10}C_r \left(1 + 3^{\frac{1}{3}}\right)^{10-7} \Rightarrow {}^{10}C_r \left(3^{\frac{1}{3}}\right)^r \) shoule be r = 0, 3
corresponding \( \left(7^{\frac{1}{7}}\right)^0 \left(3^{\frac{1}{3}}\right)^r \)
value of r = 0, 3, 6, 9 = 4 values
corresponding \( \left(7^{\frac{1}{7}}\right)^7 \left(3^{\frac{1}{3}}\right)^r \)
value of r = 0, 3 = 2 values
Total 6 values

Question. In the expansion of \( (1 + x)^n (1 + y)^n (1 + z)^n \), the sum of the co-efficients of the terms of degree 'r' is
(a) \( {}^{n^3}C_r \)
(b) \( {}^{n}C_{r^3} \)
(c) \( {}^{3n}C_r \)
(d) \( 3 \cdot {}^{2n}C_r \)
Answer: (c) \( {}^{3n}C_r \)
Solution:
Sum of the coeff of degree r is
\( (1 +x)^n (1 + y)^n (1 + z)^n \)
\( = \left( \sum_{k=0}^n {}^nC_k x^k \right) \left( \sum_{s=0}^n {}^nC_s y^s \right) \left( \sum_{t=0}^n {}^nC_t z^t \right) \)
\( = \sum_{0 \leq k,s,t \leq n} ({}^nC_k) ({}^nC_s) ({}^nC_t) x^k y^s z^t \)
degree \( m = k + s + t = r \)
sum of coeff \( = \sum_{k,s,t \geq 0} {}^nC_k \cdot {}^nC_s \cdot {}^nC_t \)
= the number of way of choosing a total number r balls out of n white, n block and n red balls.
\( = {}^{3n}C_r \)

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Question. Let \( (1 + x^2)^2 (1 + x)^n = A_0 + A_1x + A_2x^2 + \dots \). If \( A_0, A_1, A_2 \) are in A.P. then the value of n is
(a) 2
(b) 3
(c) 5
(d) 7
Answer: (a) 2, (b) 3

Question. The number \( 101^{100} - 1 \) is divisible by
(a) 100
(b) 1000
(c) 10000
(d) 100000
Answer: (a) 100, (b) 1000, (c) 10000

Question. If the \( 6^{th} \) term in the expansion of \( \left( \frac{3}{2} + \frac{x}{3} \right)^n \) when \( x = 3 \) is numerically greatest then the possible integral value(s) of n can be
(a) 11
(b) 12
(c) 13
(d) 14
Answer: (b) 12, (c) 13, (d) 14

Question. If \( (9 + \sqrt{80})^n = I + f \) where \( I, n \) are integers and \( 0 < f < 1 \), then
(a) I is an odd integer
(b) I is an even integer
(c) \( (I + f)(1 - f) = 1 \)
(d) \( 1 - f = (9 - \sqrt{80})^n \)
Answer: (a) I is an odd integer, (c) \( (I + f)(1 - f) = 1 \), (d) \( 1 - f = (9 - \sqrt{80})^n \)

Question. In the expansion of \( \left( x^{2/3} - \frac{1}{\sqrt{x}} \right)^{30} \), a term containing the power \( x^{13} \)
(a) does not exist
(b) exists & the co-efficient is divisible by 29
(c) exists & the co-efficient is divisible by 63
(d) exists & the co-efficient is divisible by 65
Answer: (b) exists & the co-efficient is divisible by 29, (c) exists & the co-efficient is divisible by 63, (d) exists & the co-efficient is divisible by 65

Question. In the expansion of \( \left( x^3 + 3 \cdot 2^{-\log_{\sqrt{2}} \sqrt{x^3}} \right)^{11} \)
(a) there appears a term with the power \( x^2 \)
(b) there does not appear a term with the power \( x^2 \)
(c) there appears a term with the power \( x^{-3} \)
(d) the ratio of the co-efficient of \( x^3 \) to that of \( x^{-3} \) is \( \frac{1}{3} \)
Answer: (b) there does not appear a term with the power \( x^2 \), (c) there appears a term with the power \( x^{-3} \), (d) the ratio of the co-efficient of \( x^3 \) to that of \( x^{-3} \) is \( \frac{1}{3} \)

Question. The co-efficient of the middle term in the expansion of \( (1 + x)^{2n} \) is
(a) \( \frac{1 \cdot 3 \cdot 5 \cdot 7 \dots (2n - 1)}{n!} 2^n \)
(b) \( ^{2n}C_n \)
(c) \( \frac{(n + 1)(n + 2)(n + 3) \dots (2n - 1)(2n)}{1 \cdot 2 \cdot 3 \dots (n - 1)n} \)
(d) \( \frac{2 \cdot 6 \cdot 10 \cdot 14 \dots (4n - 6)(4n - 2)}{1 \cdot 2 \cdot 3 \cdot 4 \dots (n - 1)n} \)
Answer: (a) \( \frac{1 \cdot 3 \cdot 5 \cdot 7 \dots (2n - 1)}{n!} 2^n \), (b) \( ^{2n}C_n \), (c) \( \frac{(n + 1)(n + 2)(n + 3) \dots (2n - 1)(2n)}{1 \cdot 2 \cdot 3 \dots (n - 1)n} \), (d) \( \frac{2 \cdot 6 \cdot 10 \cdot 14 \dots (4n - 6)(4n - 2)}{1 \cdot 2 \cdot 3 \cdot 4 \dots (n - 1)n} \)