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Question Number 54647 by gunawan last updated on 08/Feb/19

$$\mathrm{show}\mathrm{that}$$$$a.\underset{r=1}{\stackrel{n}{\mathrm{\Sigma }}}{r}^3{.}_n{C}_r=n^2(n+3).2^{n-3}$$$$b.{}_n{C}_0{.}_n{C}_1{+}_n{C}_1{.}_n{C}_2+...{+}_n{C}_{n-1}{.}_n{C}_n=\frac{\left(2n\right)!}{(n-1)!.(n+1)!}$$

Commented by maxmathsup by imad last updated on 08/Feb/19

$$lets\left(x\right)=\sum _{k=0}^n{C}_n^k{x}^k={(x+1)}^n\Rightarrow s^{{}^{\prime }}\left(x\right)=\sum _{k=1}^{n}k{C}_n^k{x}^{k-1}=n{(x+1)}^{n-1}\Rightarrow$$$$\sum _{k=1}^{n}k{C}_n^k{x}^k=nx{(x+1)}^{n-1}\Rightarrow \sum _{k=1}^n{k}^2{C}_n^k{x}^{k-1}=n{(x+1)}^{n-1}+nx(n-1){(x+1)}^{n-2}\Rightarrow$$$$\sum _{k=1}^n{k}^2{C}_n^k{x}^k=nx{(x+1)}^{n-1}+n(n-1)x^2{(x+1)}^{n-2}\Rightarrow$$$$\sum _{k=1}^n{k}^3{C}_n^k{x}^{k-1}=n{(x+1)}^{n-1}+n(n-1)x{(x+1)}^{n-2}+2n(n-1)x{(x+1)}^{n-2}$$$$+n(n-1)(n-2)x^2{(x+1)}^{n-3}\Rightarrow$$$$\sum _{k=1}^n{k}^3{C}_n^k{x}^k=nx{(x+1)}^{n-1}+n(n-1)x^2{(x+1)}^{n-2}+2n(n-1)x^2{(x+1)}^{n-2}$$$$+n(n-1)(n-2)x^3{(x+1)}^{n-3}forx=1weget$$$$\sum _{k=1}^n{k}^3{C}_n^k=n{2}^{n-1}+n(n-1)2^{n-2}+2n(n-1)2^{n-2}+n(n-1)(n-2)2^{n-3}$$$$=n{2}^{n-1}+(n^2-n+2n^2-2n)2^{n-2}+n(n-1)(n-2)2^{n-3}$$$$=n{2}^{n-1}+(3n^2-3n)2^{n-2}+n(n-1)(n-2)2^{n-3}$$$$=n{2}^{n-1}+(6n^2-6n)2^{n-3}+(n^2-n)(n-2)2^{n-3}$$$$=n{2}^{n-1}+(6n^2-6n+n^3-3n^2+2n)2^{n-3}$$$$=n{2}^{n-1}+(n^3+3n^2-4n)2^{n-3}$$$$=4n{2}^{n-3}+(n^3+3n^2-4n)2^{n-3}=(n^3+3n^2)2^{n-3}=n^2(n+3)2^{n-3}$$$$andtheresultisproved.$$$$$$

Commented by gunawan last updated on 09/Feb/19

$$\mathrm{Wow}\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}$$

Commented by Abdo msup. last updated on 09/Feb/19

$$youarewelcomesir.$$

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