Question

已知等差数列｛an｝的前n项和为Sn,且a3=5,S15=225求：设bn=3的a次方n 2n,求数列｛bn｝的前n项和Tn

Answer 1

an = a1+(n-1)d
a3=a1+2d=5 (1)
S15 = (a1+7d)15=225
a1+7d=15 (2)
(2)-(1)
d=2
a1=1
an = 1+(n-1)2 = 2n-1

bn=2n.3^an
= 2n.3^(2n-1)
consider
1+x^2+x^4+...+x^(2n)= (x^(2n+2) -1)/(x^2-1)
2x+4x^3+...+2nx^(2n-1) = [(x^(2n+2) -1)/(x^2-1)]'
=[ (x^2-1)(2n+2).x^(2n+1)- (x^(2n+2)-1)(2x)] /(x^2-1)^2
= [(2n+1)x^(2n+3)- (2n+2)x^(2n+1) + 2x]/(x^2-1)^2
put x= 3
2(3^1)+4(3^3)+...+(2n).3^(2n-1)
=(1/64)[(2n+1).3^(2n+3)- (2n+2).3^(2n+1) + 6]
=(1/64)[ 6 + (16n+7). 3^(2n+1) ]

Tn =b1+b2+...+bn
=2(3^1)+4(3^3)+...+(2n).3^(2n-1)
= (1/64)[ 6 + (16n+7). 3^(2n+1) ]