# CBSE Class 11th Math 8 – Binomial Theorem MCQs

#### (2 + 5)2 - (2 - 5)2= ____.

Correct! Wrong!

(2 + 5)5 = 5c0 25 + 5c1 245 + 5c2 2352 + 5c3 2253 + 5c4 2254 + 5c5 55 (2 - 5)5 = 5c0 25 - 5c1 245 + 5c2 2352 - 5c3 2253 + 5c4 2254 - 5c5 55 When we subtract the second expansion from the first (2 + 5)5 - (2 - 5)5 = 2(5c1 245 +5c3 2253 + 5c5 55 (2 + 5)5 - (2 - 5)5= 2(805 + 405 + 255) = 2905

#### If the coefficient of x3y8 and x8y3 in the expansion of (x + y)n is the same, then the value of n is ____.

Correct! Wrong!

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 6th and 9th terms in the expansion of (x + y)n are equal, then n is ____.

Correct! Wrong!

6th term coefficient in the given expansion is nC5. 9th term coefficient in the given expansion is nC8. Given that, their coefficients are equal. ⇒ nC5 = nC8 ⇒ n = 5 + 8 ∴ n = 13

#### The sum of the coefficients nc1 + nc2 + nc3 + ...ncn is equal to ____.

Correct! Wrong!

(1 + 1)n = nc0 + nc1 + nc2 + nc3 + ...ncn ⇒ (1 + 1)n = 1 + nc0 + nc1 + nc2 + nc3 + ...ncn 2n – 1 = nc1 + nc2 + nc3 + ...ncn

#### Number of terms in the expansion of (y2 + 2xy + x2) 30is ____.

Correct! Wrong!

(y2 + 2xy + x2) 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 ____.

Correct! Wrong!

We know that in (x + a)2 = x2 + 2ax + a2; n = 2 and the number of terms is 3. Similarly, by induction, we can prove that, (x + a)n = xn + nC1 xn-1 a + nC2 xn-2 a2 + ..... + nCr xn-r ar +...... + nCn an 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 xγ is Aγ, then Aγ - 1 + A2n - γ is ____.

Correct! Wrong!

tγ + 1 = 2n-1cγ (-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 xr−1 and x2n-γ will be equal in magnitude but opposite in sign. Then, their sum is 0, i.e. Aγ - 1 + A2n -γ = 0

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

Correct! Wrong!

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 = 15thterm rth term from the end will have (n - r) terms before it. So, it will be (n – r) + 1 from the beginning, i.e 10th term. (Here, n = 14 and r = 6)

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

Correct! Wrong!

Expansion of the first binomial gives 7 terms. Expansion of the second gives 7 terms, but here alternate terms are negative. (x + y)6 = 6c0 xn + 6c1 x5y + 6c2 x4y2 + 6c3 x3y3 + 6c4 x2y4 + 6c5 xy5 + 6c6 y6 (x - y)6 = 6c0 xn -6c1 x5y + 6c2 x4y2 -6c3 x3y3 + 6c4 x2y4 - 6c5 xy5 + 6c6 y6 When we add both expressions, the terms containing odd powers of y get cancelled. Hence, (x + y)6 + (x - y)6 = 2(6c0 x6 +6c2x4y2 +6c4x2y4 + 6c6y6)

#### The middle term in the expansion of [x3 /3 - 2/x2 ]11is ____.

Correct! Wrong!

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

#### The coefficient of x6 in the expansion (1 + x2 + x) (2 + x2)6is ____.

Correct! Wrong!

(1 + x2 + x)(2 + x2)6 = (1 + x2 + x)(26 + 6C1 25 x2 + 6C2 24 x4 + 6C3 23 x6 + ...) (The other higher powers of x are not shown). Collecting the powers of x6, 6C3 23 + 6C2 24 = 160 + 240 = 400

#### The coefficient of x5 in the expansion [1 + x + x2 + x3]13is ____.

Correct! Wrong!

[1 + x + x2 + x3]13 = (1 + x)13 [1 + x2]13 (1 + x)13 [ 1 + x2]13 = (1 + 13C1x + 13C2 x2 + 13C3 x3+ ....) (1 + 13C1 x2 + 13C2 x4 + 13C3 x6 + ...) Collecting the coefficients of x5, we get 13C1.13C2 + 13C3 .13C1 + 13C5 = 6019

#### The expression 7c0 + 7c1 + 7c2 + 7c3 + ... + 7c7is equal to ____.

Correct! Wrong!

(1 + 1)7 = 7c0 + 7c1 + 7c2 + 7c3 + ... + 7c7 = 27 = 128

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

Correct! Wrong!

In the first expansion, t n+1 = 3ncn 13n-n xn In the second expansion, tn+1 = 3n-1cn 13n-1-n xn The ratio of the coefficients = 3ncn / 3n-1cn= [3n!/2n!n! ] [(2n-1)!n!/(3n-1)!=] 3n/2n = 3/2

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

Correct! Wrong!

(1 + 7)n = nc0 1n + nc2 1n-1.7 +nc2 72 + nc3 73 + nc4 74 + ................... ncn 7n (1 + 7)n = 1 + n.7 + 72(nc2 + nc3 7+ nc4 72 + ..........ncn 7n-2) RHS is divisible by 49. Hence, LHS is also divisible by 49.

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