Question

已知数列｛an-1｝的前n项和为sn,a1＝2,2sn＝3an-4 1证明数列｛an-1｝是等比数

已知数列｛an-1｝的前n项和为sn,a1＝2,2sn＝3an-4
1证明数列｛an-1｝是等比数列
2设bn+1/3＝an+n-1/3平方n求数列｛bn｝的前n项和Mn

Answer 1

(1)
cn = an -1
Sn =b1+b2+...+bn

2Sn＝3an-4

n=1
2c1= 3a1-4
c1= (6-4)/2 = 1

2Sn＝3an-4
= 3(an -1) -1
=3bn -1
for n >=2
cn = Sn -S(n-1)
2cn = 3cn - 2c(n-1)
cn = 2c(n-1)
=> {cn} 是等比数列, q=2
=> {an - 1} 是等比数列, q=2

an - 1 = 2^(n-1) .(a1 - 1)
=2^(n-1)
an = 1+2^(n-1)

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(2)
bn+1/3＝an+n- 1/3^n
求数列｛bn｝的前n项和Mn

Solution
bn+1/3＝an+n- 1/3^n

bn+1/3＝1+2^(n-1) +n- 1/3^n
bn = 2/3 + 2^(n-1) +n- 1/3^n
Mn = b1+b2+...+bn
= (2/3)n + (2^n - 1)+ (1/2)n(n+1) - (1/2)( 1- 1/3^n)