"If $n \ge 2$ is a positive integer, then the sum of the series $${}^{n + 1}{C_2} + 2\left( {{}^2{C_2} + {}^3{C_2} + {}^4{C_2} + ... + {}^n{C_2}} \right)$$ is :"

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Accepted Answer

"$${}^{n + 1}{C_2} + 2\left( {{}^2{C_2} + {}^3{C_2} + {}^4{C_2} + ........ + {}^n{C_2}} \right)$$

$${}^{n + 1}{C_2} + 2\left( {{}^3{C_2} + {}^3{C_2} + {}^4{C_2} + ........ + {}^n{C_2}} \right)$$

use $\left\{ {{}^n{C_{r + 1}} + {}^n{C_r} = {}^{n + 1}{C_r}} \right\}$

$$ = {}^{n + 1}{C_2} + 2\left( {{}^4{C_3} + {}^4{C_2} + {}^5{C_3} + ........ + {}^n{C_2}} \right)$$

$$ = {}^{n + 1}{C_2} + 2\left( {{}^5{C_3} + {}^5{C_2} + ........ + {}^n{C_2}} \right)$$

$$\eqalign{ & . \cr & . \cr & . \cr & . \cr & . \cr} $$

$= {}^{n + 1}{C_2} + 2\left( {{}^n{C_3} + {}^n{C_2}} \right)$

$= {}^{n + 1}{C_2} + 2.{}^{n + 1}{C_3}$

$= {{(n + 1)n} \over 2} + 2.{{(n + 1)(n)(n - 1)} \over {2.3}}$

$= {{n(n + 1)(2n + 1)} \over 6}$"

If $n \ge 2$ is a positive integer, then the sum of the series $${}^{n + 1}{C_2} + 2\left( {{}^2{C_2} + {}^3{C_2} + {}^4{C_2} + ... + {}^n{C_2}} \right)$$ is :

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If $n \ge 2$ is a positive integer, then the sum of the series $${}^{n + 1}{C_2} + 2\left( {{}^2{C_2} + {}^3{C_2} + {}^4{C_2} + ... + {}^n{C_2}} \right)$$ is :

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