数列{an}满足a1=1,a2=2,an+2=(1-1/3cos^2nπ/2)an+2sin^2nπ/2 求a3,a4及数列的通项公式 设Sn=a1+a2+...an求S2n
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数列{an}满足a1=1,a2=2,an+2=(1-1/3cos^2nπ/2)an+2sin^2nπ/2 求a3,a4及数列的通项公式 设Sn=a1+a2+...an
求S2n
(1)根据 a(1)=1、a(2)=2 及 a(n+2)=[1-(1/3)cos²(nπ/2)]a(n)+2sin²(nπ/2)
可得
a(3)=[1-(1/3)cos²(π/2)]a(1)+2sin²(π/2)
=[1-(1/3)×0]×1+2×1
=3
a(4)=[1-(1/3)cos²(2π/2)]a(2)+2sin²(2π/2)
=[1-(1/3)×1]×2+2×0
=4/3
(2)由于cos、sin函数的特性,
当n为奇数时,
a(n+2)=a(n)+2;
当n为偶数时,
a(n+2)=(2/3)a(n)
可见,
a(1)、a(3)、a(5)、…、a(2n+1) 成首项为1、公差为2的等差数列,
a(2)、a(4)、a(6)、…、a(2n) 成首项为2、公差为2/3的等比数列,
所以
a(2n-1)=1+2(2n-1-1)=2n-1
a(2n)=2×[(2/3)^(n-1)]
(3)
S(2n)=1+3+5+…+(2n-1)+2×[1+(2/3)+(2/3)²+…+(2/3)^(n-1)]
=n²+2[1-(2/3)^n]/[1-(2/3)]
=n²+6[1-(2/3)^n]
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