已知等差数列{an}满足：a3=7,a5+a7=26.若bn=an+2的an-1次方,求数列｛bn｝的前n项和Sn

a5+a7=2a6=26 a6=13
a6-a3=3d=13-7=6 d=2
a1=a6-5d=13-10=3
an=a1+(n-1)d=3+2(n-1)=2n+1
bn=an +2^[a(n-1)]=2n+1+2^(2n-1)=2n+1+4^n/2
Sn=b1+b2+...+bn
=2(1+2+...+n)+ n +(1/2)(4^1+4^2+...+4ⁿ)
=2n(n+1)/2 +n +(1/2)×4×(4ⁿ-1)/(4-1)
=n²+n+n+(2/3)×(4ⁿ-1)
=n²+2n +2^(2n+1)/3 -2/3