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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)]

Findthesimplestformof∑nk=12k[sin2(2kπ3)+14]

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

wehave:sin(2kπ3)=sin(2π3)so∑nk=12k(sin2(2kπ3)+14)=∑nk=12k(sin22π3+14)=∑nk=12k((32)2+14)=∑nk=12k(34+14)=∑nk=12kletak=2k=a1qk−1=2×2k−1soq=a1=2letSn=a1+a2+...+an=a1(qn−1)q−1Sn=∑nk=02k=2(2n−1)2−1=2(2n−1)so∑nk=02k(sin2(2kπ3)+14)=2n+1−2

Commented by Joel577 last updated on 05/Oct/17

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

butifk=3,thensin2(6π3)≠sin2(2π3)

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