If n geq 2 is a positive integer, then the su | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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 ...... .

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