Question

已知等差数列{an} 的前n项和为Sn=1/2n²+3/2n+t(t∈R)

已知等差数列{an} 的前n项和为Sn=1/2n²+3/2n+t(t∈R)
(1)求数列通项an
(2)若数列{bn}满足bn=2^nan+1,求数列{bn}的前n项的和T

Answer 1

(1)
Sn =(1/2)n^2+(3/2)n +t
n=1 , a1= 2+t
an =Sn-S(n-1)
= (1/2)(2n-1) +(3/2)
= n +1
a1= 2
=>
2+t =2
t=0
ie
an =n+1

(2)
bn =2^n.an +1
= 2^n(n+1) +1
= n.2^n + 2^n +1
Tn = b1+b2+...+bn
= S + 2(2^n -1 ) + n
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)
Tn = S+ 2(2^n -1 ) + n
=n.2^(n+1) + n