已知Sn为数列｛an｝的前n项和,且a2+S2=31,a(n+1)=3an-2^n(n∈N*)求数列an的前n项和Sn

Question

已知Sn为数列｛an｝的前n项和,且a2+S2=31,a(n+1)=3an-2^n(n∈N*) 求数列an的前n项和Sn

Answer 1

a(n+1)=3an-2ⁿ
a2=3a1-2
a2+S2=a2+a1+a2=a1+2a2=a1+2(3a1-2)=7a1-4=31
7a1=35
a1=5
a(n+1)=3an -2ⁿ=3an-3×2ⁿ+2×2ⁿ=3an-3×2ⁿ+2^(n+1)
a(n+1)-2^(n+1)=3an-3×2ⁿ=3(an-2ⁿ)
[a(n+1)-2^(n+1)]/(an-2ⁿ)=3,为定值
a1-2=5-2=3,数列{an-2ⁿ}是以3为首项,3为公比的等比数列
an-2ⁿ=3ⁿ
an=2ⁿ+3ⁿ
Sn=a1+a2+...+an
=(2+2²+...+2ⁿ)+(3+3²+...+3ⁿ)
=2×(2ⁿ-1)/(2-1) +3×(3ⁿ-1)/(3-1)
=[2^(n+2) +3^(n+1)-7]/2