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

b_n =sin(a_1 +(n−1)d)⇒ S_n =?

$${b}_{{n}} ={sin}\left({a}_{\mathrm{1}} +\left({n}−\mathrm{1}\right){d}\right)\Rightarrow\:{S}_{{n}} =? \\ $$

Answered by aleks041103 last updated on 03/Nov/23

b_n =Im(e^(i(a_1 +(n−1)d)) ) S_n =Σ_(k=1) ^n b_k =Im(e^(ia_1 ) Σ_(k=1) ^n (e^(id) )^((k−1)) )= =Im(e^(ia_1 ) Σ_(k=0) ^(n−1) (e^(id) )^k )=Im(e^(ia_1 ) ((e^(ind) −1)/(e^(id) −1)))= =Im(e^(ia_1 ) ((e^(ind/2) (e^(ind/2) −e^(−ind/2) ))/(e^(id/2) (e^(id/2) −e^(−id/2) ))))= =Im(e^(i(a_1 +(n−1)d/2)) ((sin(nd/2))/(sin(d/2))))= =((sin(((nd)/2))sin(a_1 +(((n−1)d)/2)))/(sin((d/2)))) ⇒S_n =((sin(((nd)/2))sin(a_1 +(((n−1)d)/2)))/(sin((d/2))))

$${b}_{{n}} ={Im}\left({e}^{{i}\left({a}_{\mathrm{1}} +\left({n}−\mathrm{1}\right){d}\right)} \right) \\ $$$${S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{b}_{{k}} ={Im}\left({e}^{{ia}_{\mathrm{1}} } \underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({e}^{{id}} \right)^{\left({k}−\mathrm{1}\right)} \right)= \\ $$$$={Im}\left({e}^{{ia}_{\mathrm{1}} } \underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left({e}^{{id}} \right)^{{k}} \right)={Im}\left({e}^{{ia}_{\mathrm{1}} } \frac{{e}^{{ind}} −\mathrm{1}}{{e}^{{id}} −\mathrm{1}}\right)= \\ $$$$={Im}\left({e}^{{ia}_{\mathrm{1}} } \frac{{e}^{{ind}/\mathrm{2}} \left({e}^{{ind}/\mathrm{2}} −{e}^{−{ind}/\mathrm{2}} \right)}{{e}^{{id}/\mathrm{2}} \left({e}^{{id}/\mathrm{2}} −{e}^{−{id}/\mathrm{2}} \right)}\right)= \\ $$$$={Im}\left({e}^{{i}\left({a}_{\mathrm{1}} +\left({n}−\mathrm{1}\right){d}/\mathrm{2}\right)} \frac{{sin}\left({nd}/\mathrm{2}\right)}{{sin}\left({d}/\mathrm{2}\right)}\right)= \\ $$$$=\frac{{sin}\left(\frac{{nd}}{\mathrm{2}}\right){sin}\left({a}_{\mathrm{1}} +\frac{\left({n}−\mathrm{1}\right){d}}{\mathrm{2}}\right)}{{sin}\left(\frac{{d}}{\mathrm{2}}\right)} \\ $$$$ \\ $$$$\Rightarrow{S}_{{n}} =\frac{{sin}\left(\frac{{nd}}{\mathrm{2}}\right){sin}\left({a}_{\mathrm{1}} +\frac{\left({n}−\mathrm{1}\right){d}}{\mathrm{2}}\right)}{{sin}\left(\frac{{d}}{\mathrm{2}}\right)} \\ $$

Commented by hardmath last updated on 03/Nov/23

perfect solution dear professor

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

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