Question

等差数列{an}中A2=3,a4=7,数列{bn}满足bn=a下标（n）*2^n,求数列{bn}的前n项和Sn

Answer 1

设等差数列{an}公差为d.

a4-a2=2d=7-3=4

d=2

a1=a2-d=3-2=1

an=a1+(n-1)d=1+2(n-1)=2n-1

bn=an×2ⁿ=(2n-1)×2ⁿ=n×2^(n+1)-2ⁿ

令cn=n×2^(n+1)

Cn=c1+c2+...+cn=1×2²+2×2³+...+n×2^(n+1)

2Cn=1×2³+2×2⁴+...+(n-1)×2^(n+1)+n×2^(n+2)

Cn-2Cn=-Cn=2²+3³+...+2^(n+1)-n×2^(n+2)

=4×(2ⁿ-1)/(2-1) -n×2^(n+2)

=2^(n+2) -4 -n×2^(n+2)

=(1-n)×2^(n+2) -4

Cn=(n-1)×2^(n+2) +4

Sn=b1+b2+...+bn

=1×2²+2×2³+...+n×2^(n+1)-(2+2²+...+2ⁿ)

=(n-1)×2^(n+2) +4 -2×(2ⁿ-1)/(2-1)

=(2n-2)×2^(n+1) +4-2^(n+1) +2

=(2n-3)×2^(n+1) +6