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Question Number 8155 by 314159 last updated on 02/Oct/16

$$\mathrm{If}f\left(x\right)=x^n,\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}$$$$f\left(1\right)+\frac{f^1\left(1\right)}{1}+\frac{f^2\left(1\right)}{2!}+...+\frac{f^n\left(1\right)}{n!},\mathrm{where}$$$$f^r\left(x\right)\mathrm{denotes}\mathrm{the}r\mathrm{th}\mathrm{order}\mathrm{derivative}\mathrm{of}$$$$f\left(x\right)\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{is}$$

Commented by sou1618 last updated on 02/Oct/16

$$$$$$f{}^1\left(x\right)={nx}^{n-1}$$$$f{}^2\left(x\right)=n(n-1)x^{n-2}$$$$...$$$$f{}^r\left(x\right)=n(n-1)(n-2)...(n-r+1)x^{n-r}$$$$$$$$//////////$$$$findS=1+\frac{n}{1!}+\frac{n(n-1)}{2!}+\frac{n(n-1)(n-2)}{3!}+...+\frac{n!}{n!}$$$$\iff$$$$S=1+\frac{n!}{1!(n-1)!}+\frac{n!}{2!(n-2)!}+\frac{n!}{3!(n-3)!}+...+\frac{n!}{n!(n-n)!}$$$${=}_n{C}_0{+}_n{C}_1{+}_n{C}_2{+}_n{C}_3+...{+}_n{C}_n$$$$\ast {}_n{C}_r=\left(\begin{array}{c}n\\ r\end{array}\right)=\frac{n!}{r!(n-r)!}$$$$=\underset{k=0}{\sum ^n}{}_n{C}_k$$$$////////////$$$$setg\left(x\right)={(1+x)}^n$$$$g\left(x\right)=\underset{k=0}{\sum ^n}\left(1^k{x}^{n-k}{}_n{C}_k\right).$$$$g\left(1\right)=\underset{k=0}{\sum ^n}{}_n{C}_k=S$$$$g\left(1\right)={(1+1)}^n=2^n$$$$so$$$$S=2^n$$

Commented by 314159 last updated on 02/Oct/16

$$Thanks!$$

Answered by prakash jain last updated on 02/Oct/16

$$\mathrm{Answer}\mathrm{in}\mathrm{comments}$$

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