数列{an}满足a1=1,a2=2,an+2=(1-1/3cos^2nπ/2)an+2sin^2nπ/2 求a3,a4及

1个回答

  • (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]