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Question Number 199447 by jabarsing last updated on 03/Nov/23

$$b_n=sin(a_1+(n-1)d)\Rightarrow S_n=?$$

Answered by aleks041103 last updated on 03/Nov/23

$$b_n=Im\left(e^{i(a_1+(n-1)d)}\right)$$$$S_n=\underset{k=1}{\sum ^n}{b}_k=Im\left(e^{{ia}_1}\underset{k=1}{\sum ^n}{\left(e^{id}\right)}^{(k-1)}\right)=$$$$=Im\left(e^{{ia}_1}\underset{k=0}{\sum ^{n-1}}{\left(e^{id}\right)}^k\right)=Im\left(e^{{ia}_1}\frac{e^{ind}-1}{e^{id}-1}\right)=$$$$=Im\left(e^{{ia}_1}\frac{e^{ind/2}(e^{ind/2}-e^{-ind/2})}{e^{id/2}(e^{id/2}-e^{-id/2})}\right)=$$$$=Im\left(e^{i(a_1+(n-1)d/2)}\frac{sin\left(nd/2\right)}{sin\left(d/2\right)}\right)=$$$$=\frac{sin\left(\frac{nd}{2}\right)sin(a_1+\frac{(n-1)d}{2})}{sin\left(\frac{d}{2}\right)}$$$$$$$$\Rightarrow S_n=\frac{sin\left(\frac{nd}{2}\right)sin(a_1+\frac{(n-1)d}{2})}{sin\left(\frac{d}{2}\right)}$$

Commented by hardmath last updated on 03/Nov/23

$$\mathrm{perfect}\mathrm{solution}\mathrm{dear}\mathrm{professor}$$

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