# CBSE Class 11th Math 8 – Binomial Theorem MCQs

#### (2 + 5)^{2} - (2 - 5)^{2}= ____.

(2 + 5)^{5} = ^{5}c_{0} 2^{5} + ^{5}c_{1 }2^{4}5 + ^{5}c_{2} 2^{3}5^{2} + ^{5}c_{3} 2^{2}5^{3} + ^{5}c_{4} 2^{2}5^{4} + ^{5}c_{5} 5^{5 } (2 - 5)^{5} = ^{5}c_{0} 2^{5} - ^{5}c_{1 }2^{4}5 + ^{5}c_{2} 2^{3}5^{2} - ^{5}c_{3} 2^{2}5^{3} + ^{5}c_{4} 2^{2}5^{4} - ^{5}c_{5} 5^{5 } When we subtract the second expansion from the first (2 + 5)^{5} - (2 - 5)^{5} = 2(^{5}c_{1} 2^{4}5 +^{5}c_{3} 2^{2}5^{3} + ^{5}c_{5} 5^{5} (2 + 5)^{5} - (2 - 5)^{5}= 2(805 + 405 + 255) = 2905

#### If the coefficient of x^{3}y^{8} and x^{8}y^{3} in the expansion of (x + y)^{n} is the same, then the value of n is ____.

Both terms in the binomial are of degree 1. In the terms mentioned, the combined degree of x and y is 11. Therefore, n = 11

#### If the coefficients of the 6^{th} and 9^{th} terms in the expansion of (x + y)^{n } are equal, then n is ____.

6^{th }term coefficient in the given expansion is ^{n}C_{5}. 9^{th} term coefficient in the given expansion is ^{n}C_{8}. Given that, their coefficients are equal. ⇒ ^{n}C_{5} = ^{n}C_{8} ⇒ n = 5 + 8 ∴ n = 13

#### The sum of the coefficients ^{n}c_{1 }+ ^{n}c_{2 }+ ^{n}c_{3} + ...^{n}c_{n} is equal to ____.

(1 + 1)^{n} = ^{n}c_{0} + ^{n}c_{1} + ^{n}c_{2} + ^{n}c_{3} + ...^{n}c_{n } ⇒ (1 + 1)^{n} = 1 + ^{n}c_{0} + ^{n}c_{1} + ^{n}c_{2} + ^{n}c_{3} + ...^{n}c_{n } 2^{n} – 1 = ^{n}c_{1} + ^{n}c_{2} + ^{n}c_{3} + ...^{n}c_{n }

#### Number of terms in the expansion of (y^{2} + 2xy + x^{2}) ^{30}is ____.

(y^{2 }+ 2xy + x^{2}) ^{30} = [(x + y)^{2}]^{30} = (x + y)^{60} From we observed that, n = 60 Hence, the number of terms is 61.

#### If the expansion of (x + a)^{n},n ε N has 22 terms, then, n is equal to ____.

We know that in (x + a)^{2} = x^{2} + 2ax + a^{2}; n = 2 and the number of terms is 3. Similarly, by induction, we can prove that, (x + a)^{n} = x^{n} + ^{n}C_{1 }x^{n-1 }a + ^{n}C_{2 }x^{n-2 }a^{2} + ..... + ^{n}C_{r} x^{n-r }a^{r} +...... + ^{n}C_{n} a^{n} i.e. there are n + 1 terms. No. of terms = 22 ∴ n = 21

#### If in the expansion of (1- x)^{2n - 1}, the coefficient of x_{γ} is A_{γ}, then A_{γ }- 1 + A_{2n} - γ is ____.

t_{γ} + 1 = ^{2n-1}cγ (-x)^{γ } If r is odd, then r-1 is even and 2n - (r) is odd (Even number - odd number gives odd number) But, r - 1 + 2n - r = 2n -1 The coefficients of x^{r−1 }and x^{2n-γ } will be equal in magnitude but opposite in sign. Then, their sum is 0, i.e. A_{γ - 1 }+ A_{2n -γ } = 0

#### The 6^{th} term from the end, in the expansion of (x + a)^{14} is ____.

There are (n+1)^{th} terms in the expansion of (x + a)^{14}. We know that the first term from the end is the last term i.e., (n + 1)^{th} = 15^{th}term r^{th} term from the end will have (n - r) terms before it. So, it will be (n – r) + 1 from the beginning, i.e 10^{th} term. (Here, n = 14 and r = 6)

#### The number of terms in the expansion of (x+y)^{6} + (x-y)^{6}is ____.

Expansion of the first binomial gives 7 terms. Expansion of the second gives 7 terms, but here alternate terms are negative. (x + y)^{6} = ^{6}c_{0 }x^{n} + ^{6}c_{1 }x^{5}y + ^{6}c_{2} x^{4}y^{2} + ^{6}c_{3 }x^{3}y^{3} + ^{6}c_{4} x^{2}y^{4} + ^{6}c_{5 }xy^{5} + ^{6}c_{6 }y^{6} (x - y)^{6} = ^{6}c_{0 }x^{n} -^{6}c_{1 }x^{5}y + ^{6}c_{2} x^{4}y^{2} -^{6}c_{3 }x^{3}y^{3} + ^{6}c_{4} x^{2}y^{4} - ^{6}c_{5 }xy^{5} + ^{6}c_{6 }y^{6 } When we add both expressions, the terms containing odd powers of y get cancelled. Hence, (x + y)^{6} + (x - y)^{6 }= 2(^{6}c_{0} x^{6} +^{6}c_{2}x^{4}y^{2} +^{6}c_{4}x^{2}y^{4} + ^{6}c_{6}y^{6})

#### The middle term in the expansion of [x^{3 }/3 - 2/x^{2 }]^{11}is ____.

No. of terms in the expansion is 12. Therefore, there are 2 midterms, i.e. 6^{th} and 7^{th}terms.

#### The coefficient of x^{6} in the expansion (1 + x^{2} + x) (2 + x^{2})^{6}is ____.

(1 + x^{2 }+ x)(2 + x^{2})^{6} = (1 + x^{2} + x)(2^{6} + ^{6}C_{1} 2^{5} x^{2} + ^{6}C_{2} 2^{4} x^{4} + ^{6}C_{3} 2^{3} x^{6} + ...) (The other higher powers of x are not shown). Collecting the powers of x^{6}, ^{6}C_{3} 2^{3} + ^{6}C_{2} 2^{4} = 160 + 240 = 400

#### The coefficient of x^{5} in the expansion [1 + x + x^{2 }+ x^{3}]^{13}is ____.

[1 + x + x^{2} + x^{3}]^{13} = (1 + x)^{13} [1 + x^{2}]^{13} (1 + x)^{13} [ 1 + x^{2}]^{13} = (1 + ^{13}C1x + ^{13}C2 x^{2} + 13C_{3} x^{3}+ ....) (1 + ^{13}C_{1} x^{2} + ^{13}C_{2} x^{4} + ^{13}C_{3} x^{6} + ...) Collecting the coefficients of x5, we get ^{13}C_{1}.^{13}C_{2 }+ ^{13}C_{3} .^{13}C_{1} + ^{13}C_{5 }= 6019

#### The expression ^{7}c_{0} + ^{7}c_{1} + ^{7}c_{2} + ^{7}c_{3} + ... + ^{7}c_{7}is equal to ____.

(1 + 1)^{7 }= ^{7}c_{0} + ^{7}c_{1} + ^{7}c_{2} + ^{7}c_{3} + ... + ^{7}c_{7 } = 2^{7 } = 128

#### The ratio of the coefficient of x^{n} in the expansion of (1 + x)^{3n} to the coefficient of x^{n} in the expansion of (1 + x)^{3n-1}is ____.

In the first expansion, t _{n+1 }= ^{3n}c_{n} 1^{3n-n }x^{n } In the second expansion, t_{n+1 }= ^{3n-1}c_{n }1^{3n-1-n }x^{n } The ratio of the coefficients = ^{3n}c_{n} / ^{3n-1}c_{n}= [3n!/2n!n! ] [(2n-1)!n!/(3n-1)!=] 3n/2n = 3/2

#### 8^{n} - 7n - 1 is always divisible by ____.

(1 + 7)^{n} = ^{n}c_{0} 1^{n} + ^{n}c_{2} 1^{n-1}.7 +^{n}c_{2} 7^{2} + ^{n}c_{3 }7^{3} + ^{n}c_{4} 7^{4} + ................... ^{n}c_{n }7^{n} (1 + 7)^{n} = 1 + n.7 + 7^{2}(^{n}c_{2 }+ ^{n}c_{3} 7+ nc_{4} 7^{2} + ..........^{n}c_{n} 7^{n-2}) RHS is divisible by 49. Hence, LHS is also divisible by 49.