Question

What is the value of n, if the coefficients of the second term of (x – y)^3 is equal to the third term of the expansion (x + y)^n?

(a) –2

(b) 3

(c) 4

(d) 5

I have been asked this question in quiz.

My question is based upon Binomial Theorem topic in division Binomial Theorem of Mathematics – Class 11

Answer 1

The correct answer is (b) 3

Easiest explanation: Coefficient of the second term of (x – y)^3 is ^3C1 and the coefficient of the third term of the expansion (x + y)^n is ^nC2.

^3C1 = ^nC2

3 = \(\frac{n!}{2!(n – 2)!}\)

6 = \(\frac{n(n-1)(n-2)!}{(n – 2)!}\)

6 = n^2 – n

n^2 – n – 6 = 0

n^2 – 3n + 2n – 6 = 0

(n – 3) (n + 2) = 0

n = 3, – 2

Since n cannot be negative, n = 3.