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Question Number 29166 by abdo imad last updated on 04/Feb/18

$$simlify{S}_n=\sum _{k=1}^{n}k{(1+i)}^{k-1}.$$

Commented by abdo imad last updated on 06/Feb/18

$$letintroducethesequence{w}_n\left(z\right)=\sum _{k=0}^n{z}^{n}withzfromC$$$$w_n^{{}^{\prime }}\left(z\right)=\sum _{k=1}^n{nz}^{n-1}\Rightarrow S_n=w_n^{{}^{\prime }}(1+i)butwehaveforz\ne 1$$$$w_n\left(z\right)=\frac{1-z^{n+1}}{1-z}\Rightarrow w_n^{{}^{\prime }}\left(z\right)=\frac{-(n+1)z^n(1-z)-(1-z^{n+1})(-1)}{{(1-z)}^2}$$$$=\frac{-(n+1)z^n+(n+1)z^{n+1}+1-z^{n+1}}{{(1-z)}^2}$$$$=\frac{{nz}^{n+1}-(n+1)z^n+1}{{(1-z)}^2}\Rightarrow w_n^{{}^{\prime }}(1+i)=\frac{n{(1+i)}^{n+1}-(n+1){(1+i)}^n}{{(-i)}^2}$$$$=\frac{n{\left(\sqrt{2}{e}^{i\frac{\pi }{4}}\right)}^{n+1}-(n+1){\left(\sqrt{2}{e}^{i\frac{\pi }{4}}\right)}^n}{-1}$$$$=(n+1){\left(\sqrt{2}\right)}^n{e}^{i\frac{n\pi }{4}}-n{\left(\sqrt{2}\right)}^{n+1}{e}^{i\frac{(n+1)\pi }{4}}.$$$$$$

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