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Question Number 21825 by Joel577 last updated on 05/Oct/17

Find the simplest form of Σ_(k = 1) ^n 2^k [sin^2 (((2kπ)/3)) + (1/4)]

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{simplest}\:\mathrm{form}\:\mathrm{of} \\ $$$$\underset{{k}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{2}^{{k}} \left[\mathrm{sin}^{\mathrm{2}} \:\left(\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right)\:+\:\frac{\mathrm{1}}{\mathrm{4}}\right] \\ $$

Commented by sma3l2996 last updated on 05/Oct/17

we have : sin(((2kπ)/3))=sin(((2π)/3)) so Σ_(k=1) ^n 2^k (sin^2 (((2kπ)/3))+(1/4))=Σ_(k=1) ^n 2^k (sin^2 ((2π)/3)+(1/4)) =Σ_(k=1) ^n 2^k ((((√3)/2))^2 +(1/4))=Σ_(k=1) ^n 2^k ((3/4)+(1/4)) =Σ_(k=1) ^n 2^k let a_k =2^k =a_1 q^(k−1) =2×2^(k−1) so q=a_1 =2 let S_n =a_1 +a_2 +...+a_n =((a_1 (q^n −1))/(q−1)) S_n =Σ_(k=0) ^n 2^k =((2(2^n −1))/(2−1))=2(2^n −1) so Σ_(k=0) ^n 2^k (sin^2 (((2kπ)/3))+(1/4))=2^(n+1) −2

$$ \\ $$$$\:{we}\:{have}\::\:{sin}\left(\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right)={sin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right) \\ $$$${so}\:\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{2}^{{k}} \left({sin}^{\mathrm{2}} \left(\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right)+\frac{\mathrm{1}}{\mathrm{4}}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{2}^{{k}} \left({sin}^{\mathrm{2}} \frac{\mathrm{2}\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{2}^{{k}} \left(\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{2}^{{k}} \left(\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{2}^{{k}} \\ $$$${let}\:\:\:{a}_{{k}} =\mathrm{2}^{{k}} ={a}_{\mathrm{1}} {q}^{{k}−\mathrm{1}} =\mathrm{2}×\mathrm{2}^{{k}−\mathrm{1}} \\ $$$${so}\:\:{q}={a}_{\mathrm{1}} =\mathrm{2} \\ $$$${let}\:\:{S}_{{n}} ={a}_{\mathrm{1}} +{a}_{\mathrm{2}} +...+{a}_{{n}} =\frac{{a}_{\mathrm{1}} \left({q}^{{n}} −\mathrm{1}\right)}{{q}−\mathrm{1}} \\ $$$${S}_{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{2}^{{k}} =\frac{\mathrm{2}\left(\mathrm{2}^{{n}} −\mathrm{1}\right)}{\mathrm{2}−\mathrm{1}}=\mathrm{2}\left(\mathrm{2}^{{n}} −\mathrm{1}\right) \\ $$$${so} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{2}^{{k}} \left({sin}^{\mathrm{2}} \left(\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right)+\frac{\mathrm{1}}{\mathrm{4}}\right)=\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{2} \\ $$

Commented by Joel577 last updated on 05/Oct/17

but if k = 3, then sin^2 (((6π)/3)) ≠ sin^2 (((2π)/3))

$$\mathrm{but}\:\mathrm{if}\:{k}\:=\:\mathrm{3},\:\mathrm{then}\:\mathrm{sin}^{\mathrm{2}} \:\left(\frac{\mathrm{6}\pi}{\mathrm{3}}\right)\:\neq\:\mathrm{sin}^{\mathrm{2}} \:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right) \\ $$

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