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

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(B) We use the identity ${}^{n}C_{r} + {}^{n}C_{r+1} = {}^{n+1}C_{r+1}$.Note that ${}^{2}C_{2} = {}^{3}C_{3} = 1$.The series is $S = {}^{n+1}C_{2} + 2

Vedclass · Accepted Answer

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

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(B) We use the identity ${}^{n}C_{r} + {}^{n}C_{r+1} = {}^{n+1}C_{r+1}$.Note that ${}^{2}C_{2} = {}^{3}C_{3} = 1$.The series is $S = {}^{n+1}C_{2} + 2 \sum_{k=2}^{n} {}^{k}C_{2}$.Using the Hockey-stick identity $\sum_{i=r}^{n} {}^{i}C_{r} = {}^{n+1}C_{r+1}$,we have $\sum_{k=2}^{n} {}^{k}C_{2} = {}^{n+1}C_{3}$.Thus,$S = {}^{n+1}C_{2} + 2({}^{n+1}C_{3})$.$S = \frac{(n+1)n}{2} + 2 \cdot \frac{(n+1)n(n-1)}{6}$.$S = \frac{(n+1)n}{2} \left( 1 + \frac{2(n-1)}{3} \right) = \frac{n(n+1)}{2} \left( \frac{3 + 2n - 2}{3} \right)$.$S = \frac{n(n+1)(2n+1)}{6}$.

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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