Question

If n ≥ 2 is a positive integer, then the sum of the series $^{n+1}C_2 + 2( ^2C_2 + ^3C_2 + ^4C_2 + ....... + ^nC_2 )$ is :

• A. n(n-1)(2n+1)/6
• B. n(n+1)(2n+1)/6
• C. n(2n+1)(3n+1)/6
• D. n(n+1)\^2(n+2)/12

Hint:

The Hockey-stick identity $\sum_{i=r}^n {}^iC_r = {}^{n+1}C_{r+1}$ is a lifesaver for summation of binomial coefficients where the upper index varies and the lower index is constant.

Answer 1

The Correct Option is B

Step 1: Recall the Hockey-stick identity: $\sum_{i=r}^n {}^iC_r = {}^{n+1}C_{r+1}$.
Step 2: Here, $\sum_{i=2}^n {}^iC_2 = {}^{n+1}C_3$.
Step 3: The expression becomes ${}^{n+1}C_2 + 2({}^{n+1}C_3)$.
Step 4: Using ${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$, we can write: $({}^{n+1}C_2 + {}^{n+1}C_3) + {}^{n+1}C_3 = {}^{n+2}C_3 + {}^{n+1}C_3$.
Step 5: Expand: $\frac{(n+2)(n+1)n}{6} + \frac{(n+1)n(n-1)}{6} = \frac{n(n+1)}{6} [n+2 + n-1] = \frac{n(n+1)(2n+1)}{6}$. This is the sum of squares of the first $n$ natural numbers.