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

Commented by JDamian last updated on 07/Dec/19

$$Openthisapp,tapon\mathbf{Study}>\mathbf{Sequence}$$$$\mathbf{and}\mathbf{Series}andlookfor\mathbf{Sum}\mathbf{to}\mathbf{the}\mathbf{n}\mathbf{terms}$$$$\mathbf{of}\mathbf{a}\mathbf{G}.\mathbf{P}.$$

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

$$let{S}_n=\sum _{k=0}^n{e}^{ik\phi }\Rightarrow S_n=\sum _{k=0}^n{\left(e^{i\phi }\right)}^{k}if{e}^{i\phi }=1\iff \phi =2p\pi (k\in Z)$$$$S_n=(n+1)andif\phi \ne 2p\pi S_n=\frac{1-e^{i(n+1)\phi }}{1-e^{i\phi }}$$$$=\frac{1-cos(n+1)\phi -isin(n+1)\phi }{1-cos\phi -isin\phi }$$$$=\frac{2{sin}^2\left(\frac{(n+1)\phi }{2}\right)-2isin\left(\frac{(n+1)\phi }{2}\right)cos\left(\frac{(n+1)\phi }{2}\right)}{2{sin}^2\left(\frac{\phi }{2}\right)-2isin\left(\frac{\phi }{2}\right)cos\left(\frac{\phi }{2}\right)}$$$$=\frac{-isin\left(\frac{(n+1)\phi }{2}\right)e^{i(n+1)\frac{\phi }{2}}}{-isin\left(\frac{\phi }{2}\right)e^{\frac{i\phi }{2}}}=\frac{sin\left(\frac{(n+1)\phi }{2}\right)}{sin\left(\frac{\phi }{2}\right)}{e}^{in\frac{\phi }{2}}$$

Commented by chess1 last updated on 07/Dec/19

$$\mathrm{thanks}$$

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

$$youarewelcome$$

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