Question

If $n \geq 2$ is a positive integer, then the sum of the series ${ }^{n+1} C_{2}+2\left({ }^{2} C_{2}+{ }^{3} C_{2}+{ }^{4} C_{2}+\ldots .+{ }^{n} C_{2}\right)$ is:

• A. $\frac{ n ( n -1)(2 n +1)}{6}$
• B. $\frac{ n ( n +1)(2 n +1)}{6}$
• C. $\frac{ n (2 n +1)(3 n +1)}{6}$
• D. $\frac{ n ( n +1)^{2}( n +2)}{12}$

Answer 1

The Correct Option is B

To solve the given problem, we begin by understanding and simplifying the series.

The given series is:

\[{ }^{n+1} C_{2} + 2\left( { }^{2} C_{2} + { }^{3} C_{2} + { }^{4} C_{2} + \ldots + { }^{n} C_{2} \right)\]

We need to evaluate the sum: \({ }^{n+1} C_{2}\) and \(\sum_{k=2}^{n}{ }^{k} C_{2}\). Using the combination formula, where \({}^{r} C_{2} = \frac{r(r-1)}{2}\), we substitute into the series:

1. \({ }^{n+1} C_{2} = \frac{(n+1)n}{2}\)
2. For the second part: \(\sum_{k=2}^{n} { }^{k} C_{2} = \sum_{k=2}^{n} \frac{k(k-1)}{2}\)
   • Expand and simplify using summation: \(\sum_{k=2}^{n} \frac{k(k-1)}{2} = \frac{1}{2} \left[ \sum_{k=2}^{n} (k^{2} - k) \right]\)
   • Use known summation formulas:
     • \(\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\)
     • \(\sum_{k=1}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6}\)
   • Apply these to calculate:
     • \(\sum_{k=2}^{n} k = 1 + 2 + \ldots + n - (1) = \frac{n(n+1)}{2} - 1\)
     • \(\sum_{k=2}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} - 1^2\)
   • Combine and substitute back into the main expression:

Now, substituting these into the series:

\({ }^{n+1} C_{2} = \frac{(n+1)n}{2}\)
and summation terms simplify to:
\(\frac{1}{2} \left[ \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} \right]\)

Combine these in the problem's expression:

\({ }^{n+1} C_{2} + 2\left(\frac{1}{2} \left[ \sum_{k=2}^{n} (k^{2}-k) \right] \right)\)

Upon simplifying further:

\({ }^{n+1} C_{2} + \left[ \sum_{k=2}^{n} (k^{2}-k) \right]\)

Simplifying both terms gives:

The final solution becomes: \(\frac{n(n+1)(2n+1)}{6}\)

Therefore, the correct option is:

 

$\frac{ n ( n +1)(2 n +1)}{6}$