2个回答
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` n+1
a(n+1) = --------- * a(n) + 1
n-1
设 a(n+1) + m = f(n)*[a(n) + m]
` n+1 n-1
故 f(n) = ------- , m=1/(f(n)-1)= -------
` n-1 2
设 b(n) = a(n) +m
n+1 n+1 n
所以 b(n) = ------ * b(n-1) = -----*-----*b(n-2)
n-1 n-1 n-2
=...
(n+1)!/(2*3) n(n+1)
= --------------* b(2) = --------- *b(2)
(n-1)! 6
n(n+1) 2-1 n-1
即 a(n) = -----------*(a(2)+ ------) - ------
6 2 2
13 n-1
= ------ *n(n+1) - -----
12 2
1
= ----- (13n^2 +7n+6)
12
=======中学数理化解答团======
a(n+1) = --------- * a(n) + 1
n-1
设 a(n+1) + m = f(n)*[a(n) + m]
` n+1 n-1
故 f(n) = ------- , m=1/(f(n)-1)= -------
` n-1 2
设 b(n) = a(n) +m
n+1 n+1 n
所以 b(n) = ------ * b(n-1) = -----*-----*b(n-2)
n-1 n-1 n-2
=...
(n+1)!/(2*3) n(n+1)
= --------------* b(2) = --------- *b(2)
(n-1)! 6
n(n+1) 2-1 n-1
即 a(n) = -----------*(a(2)+ ------) - ------
6 2 2
13 n-1
= ------ *n(n+1) - -----
12 2
1
= ----- (13n^2 +7n+6)
12
=======中学数理化解答团======
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