# Solve this following

Question:

If $n \geq 2$ is a positive integer, then the sum of the

series ${ }^{n+1} \mathrm{C}_{2}+2\left({ }^{2} \mathrm{C}_{2}+{ }^{3} \mathrm{C}_{2}+{ }^{4} \mathrm{C}_{2}+\ldots .+{ }^{n} \mathrm{C}_{2}\right)$ is:

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

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

3. $\frac{n(2 n+1)(3 n+1)}{6}$

4. $\frac{\mathrm{n}(\mathrm{n}+1)^{2}(\mathrm{n}+2)}{12}$

Correct Option: , 2

Solution:

${ }^{\mathrm{n}+1} \mathrm{C}_{2}+2\left({ }^{2} \mathrm{C}_{2}+{ }^{3} \mathrm{C}_{2}+{ }^{4} \mathrm{C}_{2}+\ldots \ldots . .+{ }^{\mathrm{n}} \mathrm{C}_{2}\right)$

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

$\left\{\right.$ use $\left.{ }^{n} C_{r+1}+{ }^{n} C_{r}={ }^{n+1} C_{r}\right\}$

$={ }^{n+1} C_{2}+2\left({ }^{4} C_{3}+{ }^{4} C_{2}+{ }^{5} C_{3}+\ldots \ldots . .+{ }^{n} C_{2}\right)$

$={ }^{\mathrm{n}+1} \mathrm{C}_{2}+2\left({ }^{5} \mathrm{C}_{3}+{ }^{5} \mathrm{C}_{2}+\ldots \ldots+{ }^{\mathrm{n}} \mathrm{C}_{2}\right)$

$={ }^{n+1} C_{2}+2\left({ }^{n} C_{3}+{ }^{n} C_{2}\right)$

$={ }^{n+1} C_{2}+2 \cdot{ }^{n+1} C_{3}$

$=\frac{(n+1) n}{2}+2 \cdot \frac{(n+1)(n)(n-1)}{2 \cdot 3}$

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