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已知数列an满足a1=1,a3+a7=18

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已知数列an满足a1=1,a3+a7=18
 
a1=1,a3+a7=18,
a(n+1)+a(n-1)=2an
a(n+1)-an = an - a(n-1)
a(n+1)-an= a7-a6 (1)
a(n+1)-an= a6-a5 (2)
a(n+1)-an= a5-a4 (3)
a(n+1)-an= a4-a3 (4)
a(n+1)-an= a3-a2 (5)
a(n+1)-an= a2-a1 (6)
(1)+(2)+(3)+(4)+2(5)+2(6)
8[a(n+1)-an]= a7+a3- 2a1
=18 -2
a(n+1)-an=2
an -a1=2(n-1)
an = 2n-1
(2)
let
S= 1.2^1+2.2^2+...+n.2^n (1)
2S= 1.2^2+2.2^3...+n.2^(n+1) (2)
(2)-(1)
S= n.2^(n+1) -(2+2^2+...+2^n)
=n.2^(n+1) -2(2^n -1)
cn=2^(n-1).an
= 2^(n-1).( 2n-1)
= n.2^n - 2^(n-1)
Tn=c1+c2+...+cn
=S - (2^n-1)
= n.2^(n+1) -3(2^n -1)
= 3+ (2n-3).2^n