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Question Number 37890 by abdo mathsup 649 cc last updated on 19/Jun/18

$$calculate{A}_n=\sum _{k=0}^n{(-1)}^k(2k+3)intermsofn$$

Commented by prof Abdo imad last updated on 20/Jun/18

$$A_n=2\sum _{k=0}^{n}k{(-1)}^k+3\sum _{k=0}^n{(-1)}^{k}but$$$$\sum _{k=0}^n{(-1)}^k=\frac{1-{(-1)}^{n+1}}{2}$$$$letp\left(x\right)=\sum _{k=0}^n{x}^{k}wehavep\left(x\right)=\frac{x^{n+1}-1}{x-1}$$$$ifx\ne -1and{p}^{{}^{\prime }}\left(x\right)=\sum _{k=1}^{n}k{x}^{k-1}\Rightarrow$$$$x{p}^{{}^{\prime }}\left(x\right)=\sum _{k=1}^{n}k{x}^k\Rightarrow \sum _{k=1}^{n}k{(-1)}^k=-p^{{}^{\prime }}(-1)$$$$p^{{}^{\prime }}\left(x\right)=\frac{(n+1)x^n(x-1)-x^{n+1}+1}{{(x-1)}^2}$$$$=\frac{(n+1)x^{n+1}-(n+1)x^n-x^{n+1}+1}{{(x-1)}^2}$$$$=\frac{{nx}^{n+1}-(n+1)x^n+1}{{(x-1)}^2}\Rightarrow$$$$p^{{}^{\prime }}(-1)=\frac{n{(-1)}^{n+1}-(n+1){(-1)}^n+1}{4}$$$$=\frac{-n{(-1)}^n-(n+1){(-1)}^n+1}{4}$$$$=\frac{1-(2n+1){(-1)}^{\mathit{n}}}{4}\Rightarrow$$$$A_n=2\frac{(2n+1){(-1)}^n-1}{4}+\frac{3}{2}(1+{(-1)}^n)$$$$=\frac{(2n+1){(-1)}^n-1+3+3{(-1)}^n}{2}$$$$=\frac{(2n+4){(-1)}^n+2}{2}$$$$A_n=(n+2){(-1)}^n+1.$$

Commented by math khazana by abdo last updated on 20/Jun/18

$$x\ne 1$$

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