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Question Number 75122 by chess1 last updated on 07/Dec/19

Commented by JDamian last updated on 07/Dec/19

Open this app, tap on Study > Sequence and Series and look for Sum to the n terms of a G.P.

$${Open}\:{this}\:{app},\:{tap}\:{on}\:\boldsymbol{\mathrm{Study}}\:>\:\boldsymbol{\mathrm{Sequence}} \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{Series}}\:{and}\:{look}\:{for}\:\boldsymbol{\mathrm{Sum}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{n}}\:\boldsymbol{\mathrm{terms}} \\ $$$$\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{G}}.\boldsymbol{\mathrm{P}}. \\ $$

Commented by mathmax by abdo last updated on 07/Dec/19

let S_n =Σ_(k=0) ^n e^(ikϕ) ⇒ S_n =Σ_(k=0) ^n (e^(iϕ) )^k if e^(iϕ) =1 ⇔ϕ=2pπ (k∈Z) S_n =(n+1) and if ϕ≠2pπ S_n =((1−e^(i(n+1)ϕ) )/(1−e^(iϕ) )) =((1−cos(n+1)ϕ −isin(n+1)ϕ)/(1−cosϕ−isinϕ)) =((2sin^2 ((((n+1)ϕ)/2))−2isin((((n+1)ϕ)/2))cos((((n+1)ϕ)/2)))/(2sin^2 ((ϕ/2))−2isin((ϕ/2))cos((ϕ/2)))) =((−isin((((n+1)ϕ)/2)) e^(i(n+1)(ϕ/2)) )/(−isin((ϕ/2))e^((iϕ)/2) )) =((sin((((n+1)ϕ)/2)))/(sin((ϕ/2)))) e^(in(ϕ/2))

$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{e}^{{ik}\varphi} \:\:\Rightarrow\:{S}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \left({e}^{{i}\varphi} \right)^{{k}} \:\:{if}\:\:{e}^{{i}\varphi} =\mathrm{1}\:\Leftrightarrow\varphi=\mathrm{2}{p}\pi\:\:\left({k}\in{Z}\right) \\ $$$${S}_{{n}} =\left({n}+\mathrm{1}\right)\:\:{and}\:{if}\:\:\:\varphi\neq\mathrm{2}{p}\pi\:\:\:{S}_{{n}} =\frac{\mathrm{1}−{e}^{{i}\left({n}+\mathrm{1}\right)\varphi} }{\mathrm{1}−{e}^{{i}\varphi} } \\ $$$$=\frac{\mathrm{1}−{cos}\left({n}+\mathrm{1}\right)\varphi\:−{isin}\left({n}+\mathrm{1}\right)\varphi}{\mathrm{1}−{cos}\varphi−{isin}\varphi} \\ $$$$=\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\left({n}+\mathrm{1}\right)\varphi}{\mathrm{2}}\right)−\mathrm{2}{isin}\left(\frac{\left({n}+\mathrm{1}\right)\varphi}{\mathrm{2}}\right){cos}\left(\frac{\left({n}+\mathrm{1}\right)\varphi}{\mathrm{2}}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\varphi}{\mathrm{2}}\right)−\mathrm{2}{isin}\left(\frac{\varphi}{\mathrm{2}}\right){cos}\left(\frac{\varphi}{\mathrm{2}}\right)} \\ $$$$=\frac{−{isin}\left(\frac{\left({n}+\mathrm{1}\right)\varphi}{\mathrm{2}}\right)\:{e}^{{i}\left({n}+\mathrm{1}\right)\frac{\varphi}{\mathrm{2}}} }{−{isin}\left(\frac{\varphi}{\mathrm{2}}\right){e}^{\frac{{i}\varphi}{\mathrm{2}}} }\:=\frac{{sin}\left(\frac{\left({n}+\mathrm{1}\right)\varphi}{\mathrm{2}}\right)}{{sin}\left(\frac{\varphi}{\mathrm{2}}\right)}\:{e}^{{in}\frac{\varphi}{\mathrm{2}}} \\ $$

Commented by chess1 last updated on 07/Dec/19

thanks

$$\mathrm{thanks} \\ $$

Commented by mathmax by abdo last updated on 07/Dec/19

you are welcome

$${you}\:{are}\:{welcome} \\ $$

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