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Question Number 41721 by Rio Michael last updated on 11/Aug/18

$$showthat$$$$U_n=1+nx+\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^3+....$$$$forwhich{U}_n={(1+x)}^n.$$

Commented by maxmathsup by imad last updated on 12/Aug/18

$$letp\left(x\right)={(1+x)}^{n}wehavep\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{p^{\left(k\right)}\left(0\right)}{k!}{x}^k$$$$=\sum _{k=0}^n\frac{p^{\left(k\right)}\left(0\right)}{k!}{x}^k+\sum _{k=n+1}^{\mathrm{\infty }}\frac{p^{\left(k\right)}\left(0\right)}{k!}butfork>n{p}^{\left(k\right)}\left(x\right)=0\Rightarrow$$$$p\left(x\right)=\sum _{k=0}^n\frac{p^{\left(k\right)}\left(0\right)}{k!}{x}^{k}butbinomialformulaegive$$$$p\left(x\right)=\sum _{k=0}^n{C}_n^k{x}^k=C_n^0+C_n^{1}x+C_n^n{x}^2+....+C_n^n{x}^n$$$$=1+nx+\frac{n(n-1)}{2}{x}^2+.....+x^n\left(sothesumisnotinfinite\right)$$

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