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Question Number 18762 by 786786AM last updated on 29/Jul/17

$$\mathrm{If}{(1+x)}^n=C_0+C_{1}x+C_2{x}^2+...+C_n{x}^n,$$$$\mathrm{then}{C}_1{}^2+C_2{}^2+....+C_n{}^2\mathrm{is}\mathrm{equal}\mathrm{to}$$

Answered by Rishabh#1 last updated on 30/Jul/17

$${(1+x)}^{2n}={(1+x)}^n{(x+1)}^n$$$$=({}^n{C}_0{x}^0+{}^n{C}_{1}x+...+{}^n{C}_n{x}^n)\times$$$$({}^n{C}_0{x}^n+{}^n{C}_1{x}^{n-1}+...{+}^n{C}_n{x}^0)$$$$coeffecientof{x}^{n}inlhs={}^{2n}{C}_n$$$$coeffecientof{x}^{n}inrhs=$$$$C_0^2+C_1^2+..+C_n^2$$$$hence$$$$C_0^2+C_1^2+..+C_n^2{=}^{2n}{C}_n$$$$$$

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