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

simlify S_n = Σ_(k=1) ^n  k(1+i)^(k−1)     .

$${simlify}\:{S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}\left(\mathrm{1}+{i}\right)^{{k}−\mathrm{1}} \:\:\:\:. \\ $$

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

let introduce the sequence w_n (z)=Σ_(k=0) ^n  z^n    with z from C  w_n ^′ (z)= Σ_(k=1) ^n  nz^(n−1)   ⇒ S_n =w_n ^′ (1+i) butwe havefor z≠1  w_n (z)=((1−z^(n+1) )/(1−z))⇒w_n ^′ (z)=((−(n+1)z^n (1−z)−(1−z^(n+1) )(−1))/((1−z)^2 ))  =((−(n+1)z^n  +(n+1)z^(n+1)  +1 −z^(n+1) )/((1−z)^2 ))  =((nz^(n+1)  −(n+1)z^n  +1)/((1−z)^2 ))⇒w_n ^′ (1+i)=((n(1+i)^(n+1) −(n+1)(1+i)^n )/((−i)^2 ))  =((n((√2)e^(i(π/4)) )^(n+1)  −(n+1)((√2) e^(i(π/4)) )^n )/(−1))  =(n+1)((√2))^n  e^(i((nπ)/4))  −n ((√2))^(n+1)  e^(i(((n+1)π)/4))    .

$${let}\:{introduce}\:{the}\:{sequence}\:{w}_{{n}} \left({z}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{z}^{{n}} \:\:\:{with}\:{z}\:{from}\:{C} \\ $$$${w}_{{n}} ^{'} \left({z}\right)=\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{nz}^{{n}−\mathrm{1}} \:\:\Rightarrow\:{S}_{{n}} ={w}_{{n}} ^{'} \left(\mathrm{1}+{i}\right)\:{butwe}\:{havefor}\:{z}\neq\mathrm{1} \\ $$$${w}_{{n}} \left({z}\right)=\frac{\mathrm{1}−{z}^{{n}+\mathrm{1}} }{\mathrm{1}−{z}}\Rightarrow{w}_{{n}} ^{'} \left({z}\right)=\frac{−\left({n}+\mathrm{1}\right){z}^{{n}} \left(\mathrm{1}−{z}\right)−\left(\mathrm{1}−{z}^{{n}+\mathrm{1}} \right)\left(−\mathrm{1}\right)}{\left(\mathrm{1}−{z}\right)^{\mathrm{2}} } \\ $$$$=\frac{−\left({n}+\mathrm{1}\right){z}^{{n}} \:+\left({n}+\mathrm{1}\right){z}^{{n}+\mathrm{1}} \:+\mathrm{1}\:−{z}^{{n}+\mathrm{1}} }{\left(\mathrm{1}−{z}\right)^{\mathrm{2}} } \\ $$$$=\frac{{nz}^{{n}+\mathrm{1}} \:−\left({n}+\mathrm{1}\right){z}^{{n}} \:+\mathrm{1}}{\left(\mathrm{1}−{z}\right)^{\mathrm{2}} }\Rightarrow{w}_{{n}} ^{'} \left(\mathrm{1}+{i}\right)=\frac{{n}\left(\mathrm{1}+{i}\right)^{{n}+\mathrm{1}} −\left({n}+\mathrm{1}\right)\left(\mathrm{1}+{i}\right)^{{n}} }{\left(−{i}\right)^{\mathrm{2}} } \\ $$$$=\frac{{n}\left(\sqrt{\mathrm{2}}{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)^{{n}+\mathrm{1}} \:−\left({n}+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)^{{n}} }{−\mathrm{1}} \\ $$$$=\left({n}+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}\right)^{{n}} \:{e}^{{i}\frac{{n}\pi}{\mathrm{4}}} \:−{n}\:\left(\sqrt{\mathrm{2}}\right)^{{n}+\mathrm{1}} \:{e}^{{i}\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{4}}} \:\:\:. \\ $$$$ \\ $$

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