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

${ }^{n+1} C_{2}+2\left({ }^{2} C_{2}+{ }^{3} C_{2}+{ }^{4} C_{2}+\ldots \ldots .+{ }^{n} C_{2}\right)$ ${ }^{n+1} C_{2}+2\left({ }^{3} C_{3}+{ }^{3} C_{2}+{ }^{4} C_{2}+\ldots \ldots .+{ }^{n} C_{2}\right)$ $\left\{\right.$ use $\left.{ }^{n} C_{r+1}+{ }^{n} C_{r}={ }^{n+1} C_{r}\right\}$ $={ }^{n+1} C_{2}+2\left({ }^{4} C_{3}+{ }^{4} C_{2}+{ }^{5} C_{3}+\ldots \ldots+{ }^{n} C_{2}\right)$ ${ }^{n+1} C_{2}+2\left({ }^{5} C_{3}+{ }^{5} C_{2}+\ldots \ldots .+{ }^{n} C_{2}\right) \vdots$ $={ }^{n+1} C_{2}+2\left({ }^{n} C_{3}+{ }^{n} C_{2}\right)$ $={ }^{n+1} C_{2}+2 \cdot{ }^{n+1} C_{3}$ $=\frac{(n+1) n}{2}+2 \cdot \frac{(n+1)(n)(n-1)}{2.3}$ $ =\frac{n(n+1)(2 n+1)}{6}$