Question

等比数列{a_n}的首项a₁=，且4a_n-1+a_n+1=4a_n,则sina₁+sina₂+sina₃+…+sina₂₀₁₄=

Answer 1

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设等比数列的公比为q
∵4a_n-1+a_n+1=4a_n
∴4a_n-1+a_n-1q²=4a_n-1q
∴4+q²=4qq=2a_n=·2^n-1
∴a₁=, a₂=·2=, a₃=·2²=,a₄=·2³=,a₅=·2⁴=, a₆=·2⁴=,…
∴sina₁=sin=,sina₂=sin=,
sina₃=sin= sin()=,
sina₄=sin(·2²)=sin=sin()=-sin=-
sina₅=sin(·2³)=sin=sin()=sin= sin()=
sina₆=sin(·2⁴)=sin= sin()=sin()=-sin=-
一般地,当n≥3时,设n=2k+1(k=1,2,3,…),则
a_n=·2^n-1=·2^2k-2=·4^k-1=·(1+3)^k-1=·(1+3+3²+…+3^k-1)
=+(+3¹+…+3^k-2)(规定=0)
∴sina_n=sin[+(+3¹+…+3^k-2)]=sin= sin()=,
设n=2k+2(k=1,2,3,…),则
a_n=·2^n-1=·2^2k-2=·4^k-1=·(1+3)^k-1=·(1+3+3²+…+3^k-1)
=+(+3¹+…+3^k-2)
∴sina_n=sin[+(+3¹+…+3^k-2)]= sin=sin()=-sin=-
所以,当n≥3时,sina_2k+1+sina_2k+2=0(k=1,2,3,…)
∴sina₁+sina₂+sina₃+…+sina₂₀₁₄= sina₁+sina₂=+