Question

已知数列an=2n+1 Sn=n^2+2n bn=2^n b1=2 cn=an*bn 求cn的前n项和Tn

Answer 1

an=2n+1 bn=2ⁿ
cn=anbn=(2n+1)×2ⁿ=n×2^(n+1) +2ⁿ
Tn=c1+c2+...+cn
=1×2²+2×2³+...+n×2^(n+1) +(2+2²+...+2ⁿ)
令Cn=1×2²+2×2³+3×2⁴+...+n×2^(n+1)
则2Cn=1×2³+2×2⁴+...+(n-1)×2^(n+1)+n×2^(n+2)
Cn-2Cn=-Cn=2²+2³+...+2^(n+1)-n×2^(n+2)
Cn=n×2^(n+2)-[2²+2³+...+2^(n+1)]
Tn=Cn+(2+2²+...+2ⁿ)
=n×2^(n+2)-[2²+2³+...+2^(n+1)]+(2+2²+...+2ⁿ)
=n×2^(n+2) +2-2^(n+1)