作业帮求数列递推公式
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求数列递推公式
求数列递推公式
求数列递推公式
b(n) = [2a(n+1)-a(n)]/4,
b(n+1) = [2a(n+2)-a(n+1)]/4.
2^(3/2)b(n+1) = 2^(3/2)[2a(n+2)-a(n+1)]/4 = b(n) + 2^(1/2)a(n) = [2a(n+1)-a(n)]/4 + 2^(1/2)a(n),
2a(n+2)-a(n+1) = 2^(1/2)[2a(n+1)-a(n)]/4 + 2a(n)
= a(n+1)/2^(1/2) - a(n)/2^(3/2) + 2a(n),
2a(n+2) = [1+2^(-1/2)]a(n+1) + [2-2^(-3/2)]a(n).
a(n+2) = [1/2 + 2^(-3/2)]a(n+1) + [1 - 2^(-5/2)]a(n).
a(n) = [2^(3/2)b(n+1) - b(n)]/2^(1/2).
a(n+1) = [2^(3/2)b(n+2) - b(n+1)]/2^(1/2).
2a(n+1) = 2[2^(3/2)b(n+2) - b(n+1)]/2^(1/2) = a(n) + 4b(n) = [2^(3/2)b(n+1) - b(n)]/2^(1/2) + 4b(n)
4b(n+2) - 2^(1/2)b(n+1) = 2b(n+1) - b(n)/2^(1/2) + 4b(n),
4b(n+2) = [2 + 2^(1/2)]b(n+1) + [4-1/2^(1/2)]b(n),
b(n+2) = [1/2 + 2^(-3/2)]b(n+1) + [1 - 2^(-5/2)]b(n).
综合,有,
a(n+2) = [1/2 + 2^(-3/2)]a(n+1) + [1-2^(-5/2)]a(n),
b(n+2) = [1/2 + 2^(-3/2)]b(n+1) + [1-2^(-5/2)]b(n).