Question

已知数列{an}满足an=2an-1+2^n-1(n∈N+,且n≥2),a4=81.求数列{an}的前n项和Sn.用简单

Answer 1

an=2a(n-1)+2^n-1

an-1=2(a(n-1)-1)+2^n

令bn=an-1

bn=2b(n-1)+2^n

bn/2^n=b(n-1)/2^(n-1)+1

bn/2^n=b1/2+(n-1)

bn=((a1-1)/2+n-1)*2^n

an=((a1-1)/2+n-1)*2^n+1

a4=((a1-1)/2+3)*16+1

(a1-1)/2+3=5

a1=5

an=(n+1)*2^n+1

令cn=(n+1)*2^n

cn前n项和为tn,则sn=n+tn

tn=2*2+3*2^2+...+(n+1)*2^n

2tn=2*2^2+3*2^3+...+(n+1)*2^(n+1)

2tn-tn=(n+1)*2^(n+1)+n*2^n-(n+1)*2^n+...+2*2^2-3*2^2-2*2

tn=(n+1)*2^(n+1)-2^n-...-2^2-2-1-1

tn=(n+1)*2^(n+1)-(1+2+2^2+...+2^n)-1

tn=(n+1)*2^(n+1)-2^(n+1)

tn=n*2^(n+1)

sn=n+n*2^(n+1)