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Question Number 37891 by abdo mathsup 649 cc last updated on 19/Jun/18

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

Commented by prof Abdo imad last updated on 24/Jun/18

$$B_n=2\sum _{k=0}^n{k}^2{(-1)}^k+\sum _{k=0}^n{(-1)}^k$$$$letp\left(x\right)=\sum _{k=0}^n{x}^{k}withx\ne 1$$$$p^{{}^{\prime }}\left(x\right)=\sum _{k=1}^{n}k{x}^{k-1}\Rightarrow x{p}^{{}^{\prime }}\left(x\right)=\sum _{k=1}^n{kx}^k\Rightarrow$$$$\sum _{k=1}^n{k}^2{x}^{k-1}=p^{{}^{\prime }}\left(x\right)+{xp}^{″}\left(x\right)\Rightarrow$$$$\sum _{k=1}^n{k}^2{x}^k={xp}^{{}^{\prime }}\left(x\right)+x^2{p}^{″}\left(x\right)but$$$$p\left(x\right)=\frac{x^{n+1}-1}{x-1}\Rightarrow p^{{}^{\prime }}\left(x\right)=\frac{n{x}^{n+1}-(n+1)x^n+1}{{(x-1)}^2}$$$$p^{{}^{″}}\left(x\right)=\frac{(n(n+1)x^n-n(n+1)x^{n-1}){(x-1)}^2-2(x-1)({nx}^{n+1}-(n+1)x^n+1)}{{(x-1)}^4}$$$$=\frac{(x-1)n(n+1)(x^n-x^{n-1})-2({nx}^{n+1}-(n+1)x^n+1)}{{(x-1)}^3}$$$$\sum _{k=1}^n{k}^2{(-1)}^k=-p^{{}^{\prime }}(-1)+p^{{}^{″}}(-1)$$$$=-\frac{n{(-1)}^{n+1}-(n+1){(-1)}^n+1}{4}+\frac{-2n(n+1)({(-1)}^n-{(-1)}^{n-1})-2\{n{(-1)}^{n+1}-(n+1){(-1)}^n+1\}}{-8}$$$$also\sum _{k=0}^n{(-1)}^k=\frac{1-{(-1)}^{n+1}}{2}$$$$sothevalueofSisdetermined.$$

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