$^{n + 1} C_{2} + 2 [^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + . . . +^{n} C_{2}] =$

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

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

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\frac{n (n - 1) (2 n - 1)}{6}

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None of these

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Solution

The correct option is A $\frac{n (n + 1) (2 n + 1)}{6}$
We have,
$^{n + 1} C_{2} + 2 [^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + . . . +^{n} C_{2}] =^{n + 1} C_{2} + 2 [^{3} C_{3} +^{3} C_{2} +^{4} C_{2} + . . . +^{n} C_{2}] =^{n + 1} C_{2} + 2 [^{4} C_{3} +^{4} C_{2} + . . . +^{n} C_{2}] =^{n + 1} C_{2} + 2 [^{5} C_{3} + . . . +^{n} C_{2}] =^{n + 1} C_{2} + {2.}^{n + 1} C_{3} =^{n + 1} C_{2} +^{n + 1} C_{3} +^{n + 1} C_{3} =^{n + 2} C_{3} +^{n + 1} C_{3}$
$= \frac{n (n + 1) (n + 2)}{6} + \frac{n (n + 1) (n - 1)}{6} = \frac{n (n + 1) (2 n + 1)}{6}$

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