已知数列{an}中,a1=-5/8,an+1-an=1/n*(n+1)(n属于N,求a1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的最小值?

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已知数列{an}中,a1=-5/8,an+1-an=1/n*(n+1)(n属于N,求a1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=(

已知数列{an}中,a1=-5/8,an+1-an=1/n*(n+1)(n属于N,求a1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的最小值?
已知数列{an}中,a1=-5/8,an+1-an=1/n*(n+1)(n属于N,求a1
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的最小值?

已知数列{an}中,a1=-5/8,an+1-an=1/n*(n+1)(n属于N,求a1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的最小值?
(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