Question

If the coefficients of x^r–1, x^r, x^r+1 in the binomial expansion of (1 + x)ⁿ are in A.P., then n² – kn + 4r² – 2 = 0, where k =

• A. r + 1
• B. 2r + 1
• C. 4r + 1
• D. None of these

Answer 1

Answer: (C)

4r + 1

Answer 2

The co-efficients of x^r–1, x^r, x^r+1 in the expansion of (1 + x)ⁿ are respectively ⁿC_r–1, ⁿC_r, ⁿC_r+1^, which are in A.P.

\ 2 · ⁿC_r = ⁿC_r–1 + ⁿC_r+1

or 2 · $\frac{n!}{r!(n - r)!}$=$\frac{n!}{(r - 1)!(n - r + 1)!}$+ $\frac{n!}{(r + 1)!(n - r - 1)!}$

or $\frac{2}{r(r - 1)!(n - r)(n - r - 1)!}$

=$\frac{1}{(r - 1)!(n - r + 1)(n - r)(n - r - 1)!}$+$\frac{1}{(r + 1)r(r - 1)!(n - r - 1)!}$

or $\frac{2}{r(n - r)}$= $\frac{1}{(n - r + 1)(n - r)}$+$\frac{1}{r(r + 1)}$

or 2(n– r + 1) (r + 1) = r (r + 1) + (n – r + 1) (n – r)

or n² – n (4r + 1) + 4r² – 2 = 0.