Question

已知数列{an}满足a1=2,a(n+1)=2an+3

(1)求｛an｝的通项公式
(2)求数列｛nan｝的前n项和Sn

Answer 1

(1)
a(n+1)=2an+3
a(n+1)+3 = 2(an+3)
[a(n+1)+3]/[an+3] = 2
(an+3)/(a1+3) = 2^(n-1)
an= 5.2^(n-1) -3
(2)
nan = 5[n.2^(n-1)] - 3n
Sn = summation ( 5[n.2^(n-1)] - 3n)
consider
1+x+x^2+..+x^n = [x^(n+1) - 1]/(x-1)
1+2x+3x^2+..+nx^(n-1)
={[x^(n+1) - 1]/(x-1)}'
=[nx^(n+1) - (n+1)x^n + 1] /(x-1)^2
put x = 2
1.2^0 +2.2^1+...+n.2^(n-1)
=n.2^(n+1) - (n+1)2^n + 1

Sn = summation ( 5[n.2^(n-1)] - 3n)
= 5[n.2^(n+1) - (n+1)2^n + 1] - 3n(n+1)/2