Question

数学/-*/*-/*-/-*/-*/*-*/++++++++++++++++++++++++

设数列｛an｝的前n项和为Sn,若对任意的正整数n都有Sn=2an-3n。
（1）设bn=an+3，求证：数列｛bn｝是等比数列。并求出｛an｝的通项公式
（2）求出数列｛｝的前n项和

Answer 1

Sn=2an-3n
Sn-1=2an-1-3n+3
Sn-Sn-1=an
an=2an-2an-1-3
an+3=2(an-1+3)
bn=an+3
所以bn为等比数列
a1=3,a2=9,a3=21
b1=6,b2=12,b3=24
bn=3*2^n
an+3=3*2^n
an=3*2^n-3

nan=3n*2^n-3n
前n项和为Sn
Sn=3(1*2^1+2*2^2+3*2^3+4*2^4+5*2^5+....n*2^n)-3n(n+1)/2
2Sn=3(1*2^2+2*2^3+3*2^4+4*2^5+5*2^6+....n*2^n+1)-3n(n+1)
-Sn=3(2^1+2^2+2^3+2^4+.....2^n-n*2^n+1)+3n(n+1)/2
=3(2^n+1-2-n*2^n+1)+3n(n+1)/2
Sn=3(n-1)2^n+1-3n(n+1)/2-6