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Question Number 185982 by Shrinava last updated on 30/Jan/23

Answered by Mathspace last updated on 30/Jan/23

$$wedeveloppfirst{(x+1)}^{2n}$$$$=\sum _{k=0}^{2n}{C}_{2n}^k{x}^{k}but$$$${(x+1)}^{2n}={(x+1)}^n.{(x+1)}^n$$$$=\left(\sum _{k=0}^n{C}_n^k{x}^k\right)\left(\sum _{k=0}^n{C}_n^k{x}^k\right)$$$$=\sum _{k=0}^{2n}{a}_k{x}^k$$$$a_k=\sum _{i+j=k}{a}_i{a}_j$$$$=\sum _{i=0}^k{a}_i{a}_{k-i}=\sum _{i=0}^k{C}_k^i{C}_k^{k-i}$$$$=\sum _{i=0}^k{\left(C_k^i\right)}^2$$$$theequalitygive$$$$\sum _{i=0}^k{\left(C_k^i\right)}^2=C_{2n}^k$$$$k=n\Rightarrow \sum _{i=0}^n{\left(C_n^i\right)}^2=C_{2n}^n$$$$=\frac{\left(2n\right)!}{\left(n\right){!}^2}$$

Commented by Shrinava last updated on 30/Jan/23

$$\mathrm{cool}\mathrm{dear}\mathrm{professor}\mathrm{thank}\mathrm{you}$$

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