已知数列an中,a1=-5/8,an+1-an=1/n(n+1) .(1)求a2,a3 (2)求an (3)设bn=(1+2+3+...+n）an,求bn最小值急啊

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已知数列an中,a1=-5/8,an+1-an=1/n(n+1) .(1)求a2,a3 (2)求an (3)设bn=(1+2+3+...+n）an,求bn最小值急啊
已知数列an中,a1=-5/8,an+1-an=1/n(n+1) .(1)求a2,a3 (2)求an (3)设bn=(1+2+3+...+n）an,求bn最小值
急啊

已知数列an中,a1=-5/8,an+1-an=1/n(n+1) .(1)求a2,a3 (2)求an (3)设bn=(1+2+3+...+n）an,求bn最小值急啊
a(n+1) - a(n) = 1/n - 1/(n+1),
a(n+1) + 1/(n+1) = a(n) + 1/n.
{a(n)+1/n}是首项为a(1)+1 = 3/8的常数数列.
a(n)+1/n = 3/8,
a(n) = 3/8 - 1/n.
a(2) = 3/8 - 1/2 = -1/8.
a(3) = 3/8 - 1/3 = 1/24.
b(n) = [1+2+...+n]a(n) = n(n+1)/2[3/8 - 1/n] = (n+1)[3n/16 - 1/2]
= (n+1)(3n - 8)/16
= (3n^2 - 5n - 8)/16
= (3/16)[n^2 - 5n/3 - 8/3]
= (3/16)[n^2 - 2*5n/6 + 25/36 - 8/3 - 25/36]
= (3/16)[(n-5/6)^2 - 121/36],
n>=1时,b(n)单调递增.
b(n) >= b(1) = 2*(-5)/16 = -5/8.

(1) a2=-1/8，a3=1/24
(2)a2-a1=1-1/2
a3-a2=1/2-1/3
a4-a3=1/3-1/4
... ...
an-an-1=1/(n-1)-1/n
累加得 an-a1=1-1/n
an=3/8-1/n
（3）b...

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(1) a2=-1/8，a3=1/24
(2)a2-a1=1-1/2
a3-a2=1/2-1/3
a4-a3=1/3-1/4
... ...
an-an-1=1/(n-1)-1/n
累加得 an-a1=1-1/n
an=3/8-1/n
（3）bn=n(n+1)/2*(3n-1/8n)
= (3n-1)(n+1)/4
=3/4×（n^2+2/3n-1/3）
=3/4*[(n+1/3)^2-4/9]
所以 bn在（1，+∞）上递增
bn min=b1=-5/8

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