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:
Correct Option: , 2
${ }^{\mathrm{n}+1} \mathrm{C}_{2}+2\left({ }^{2} \mathrm{C}_{2}+{ }^{3} \mathrm{C}_{2}+{ }^{4} \mathrm{C}_{2}+\ldots \ldots . .+{ }^{\mathrm{n}} \mathrm{C}_{2}\right)$
${ }^{n+1} C_{2}+2\left({ }^{3} C_{3}+{ }^{3} C_{2}+{ }^{4} C_{2}+\ldots \ldots . .+{ }^{n} C_{2}\right)$
$\left\{\right.$ use $\left.{ }^{n} C_{r+1}+{ }^{n} C_{r}={ }^{n+1} C_{r}\right\}$
$={ }^{n+1} C_{2}+2\left({ }^{4} C_{3}+{ }^{4} C_{2}+{ }^{5} C_{3}+\ldots \ldots . .+{ }^{n} C_{2}\right)$
$={ }^{\mathrm{n}+1} \mathrm{C}_{2}+2\left({ }^{5} \mathrm{C}_{3}+{ }^{5} \mathrm{C}_{2}+\ldots \ldots+{ }^{\mathrm{n}} \mathrm{C}_{2}\right)$
$={ }^{n+1} C_{2}+2\left({ }^{n} C_{3}+{ }^{n} C_{2}\right)$
$={ }^{n+1} C_{2}+2 \cdot{ }^{n+1} C_{3}$
$=\frac{(n+1) n}{2}+2 \cdot \frac{(n+1)(n)(n-1)}{2 \cdot 3}$
$=\frac{n(n+1)(2 n+1)}{6}$
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