Question

设sn为数列（an）的前n项和，已知a1≠0，2an一a1=s1xsn，n∈N，求a1，a2，和（

设sn为数列（an）的前n项和，已知a1≠0，2an一a1=s1xsn，n∈N，求a1，a2，和（an）的通项公式，和（na）的前n项和，

Answer 1

2an-a1 = S1.Sn
n=1
a1= (a1)^2
a1 =1

n=2
2a2- a1 = a1.(a1+a2)
2a2-1 = 1+a2
a2= 2

2an-a1 = S1.Sn
2[Sn-S(n-1)] -1 = Sn
Sn +1 = 2[S(n-1) + 1]
{Sn +1 } 是等比数列, q=2
Sn +1 = 2^(n-1).( S1 +1 )
= 2^n
Sn = -1+2^n
an = Sn -S(n-1)
= 2^(n-1)

bn = nan
= n.2^(n-1)
let
S = 1.2^0+ 2.2^1+....+n.2^(n-1) (1)
2S = 1.2^1+ 2.2^2+....+n.2^n (2)
(2)-(1)
S = n.2^n - [1+2+2^2+...+2^(n-1)]
= n.2^n - (2^n -1)
= 1+ (n-1).2^n
Tn = b1+b2+...+bn
= S
= 1+ (n-1).2^n