If n > 2 is a positive integer, then the sum of the series

Question:

If $n \geq 2$ is a positive integer, then the sum of the series ${ }^{\mathrm{n}+1} \mathrm{C}_{2}+2\left({ }^{2} \mathrm{C}_{2}+{ }^{3} \mathrm{C}_{2}+{ }^{4} \mathrm{C}_{2}+\ldots+{ }^{\mathrm{n}} \mathrm{C}_{2}\right)$ is :

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

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

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

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


Correct Option: , 3

Solution:

${ }^{2} C_{2}={ }^{3} C_{3}$

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

$\because{ }^{n} C_{r}+{ }^{n} C_{r-1}={ }^{n+1} C_{r}$

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

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

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

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