Question

在数列{an}中，已知Sn为数列{an}的前n项和，且Sn/（an-n）=1+n （1）求数列{1

在数列{an}中，已知Sn为数列{an}的前n项和，且Sn/（an-n）=1+n （1）求数列{1/Sn}的前2014项和M （2）求数列{an•3^n}的前n项和Tn

Answer 1

(1)
Sn/（an-n）=1+n
n=1,
a1/(a1-1) = 2
a1=2a1-2
a1= 2

Sn = (1+n)(an-n) (1)
S(n-1) = n[a(n-1)-(n-1)] (2)
(1)-(2)
an =(1+n)(an-n) -n[a(n-1)-(n-1)]
nan = n(1+n)-n(n-1) +na(n-1)
an-a(n-1) = 2
an -a1 =2(n-1)
an=2n

Sn = n(n+1)
1/Sn = 1/n -1/(n+1)
Gn = 1/S1+1/S2+...+1/Sn
= 1-1/(n+1)
= n/(n+1)
G2014= 2014/2015
{1/Sn}的前2014项和=G2014=2014/2015

(2)
let
S= 1.3^1+2.3^2+...+n.3^n (1)
3S= 1.3^2+2.3^3+...+n.3^(n+1) (2)
(2)-(1)
2S = n.3^(n+1) - (3+3^2+....+3^n)
=n.3^(n+1) - (3/2)(3^n-1)
=(3/2) + (3n -3/2).3^n

bn = an.3^n
= 2(n.3^n)
Tn=b1+b2+...+bn
= 2S
=(3/2) + (3n -3/2).3^n

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