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Question Number 25969 by soyebshaikh41@gmail.com last updated on 17/Dec/17

$${\mathrm{C}}_0+{2\mathrm{C}}_1+{3\mathrm{C}}_2+..........+(\mathrm{n}+1){\mathrm{C}}_{\mathrm{n}}=(\mathrm{n}+2)2^{\mathrm{n}-1}$$$$\mathrm{using}\mathrm{Bionomial}\mathrm{teorm}$$

Commented by moxhix last updated on 17/Dec/17

$$show{}_n{C}_0+2{}_n{C}_1+3{}_n{C}_2+...+(n+1){}_n{C}_n=(n+2)2^{n-1}$$$$putf\left(x\right)=x{(1+x)}^n$$$$$$$$f{}^{\prime }\left(x\right)={(1+x)}^n+nx{(1+x)}^{n-1}=(n+1+x){(1+x)}^{n-1}$$$$f{}^{\prime }\left(1\right)=(n+2)2^{n-1}$$$$$$$$f\left(x\right)=x\underset{k=0}{\sum ^n}{}_n{C}_k{x}^k=\underset{k=0}{\sum ^n}{}_n{C}_k{x}^{k+1}$$$$f{}^{\prime }\left(x\right)=\underset{k=0}{\sum ^n}{}_n{C}_k(k+1)x^k$$$$f{}^{\prime }\left(1\right)=\underset{k=0}{\sum ^n}{}_n{C}_k(k+1)$$$$$$$$\therefore {}_n{C}_0+2{}_n{C}_1+3{}_n{C}_2+...+(n+1){}_n{C}_n=(n+2)2^{n-1}$$

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