已知a1=1,2an+1*an+3an+1+an+2=0 求证{1/an+1}为等差数列(2)求an
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已知a1=1,2an+1*an+3an+1+an+2=0 求证{1/an+1}为等差数列(2)求an
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(1)
2a(n+1)*a(n) + 3a(n+1) + a(n) + 2 = 0
2a(n+1)*a(n) + 2a(n+1) + 2a(n) + 2 = a(n) - a(n+1)
2[a(n+1) +1]*[a(n) + 1] = [a(n) + 1] - [a(n+1) + 1]
2 = 1/[a(n) + 1] - 1/[a(n+1) + 1]
显然 { 1/[a(n) + 1] } 是首项为 1/2,公差为-2的等差数列.
(2) 由(1)可得a(n) = -2n - 0.5
(1)
2a(n+1)*a(n) + 3a(n+1) + a(n) + 2 = 0
2a(n+1)*a(n) + 2a(n+1) + 2a(n) + 2 = a(n) - a(n+1)
2[a(n+1) +1]*[a(n) + 1] = [a(n) + 1] - [a(n+1) + 1]
2 = 1/[a(n) + 1] - 1/[a(n+1) + 1]
显然 { 1/[a(n) + 1] } 是首项为 1/2,公差为-2的等差数列.
(2) 由(1)可得a(n) = -2n - 0.5
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