An=(n*2^n-2^n+1)⼀ [(n+1)*(n^2+2^n)] 求Sn大神求解啊急呢在线等

a(n+1)=2^(n+1)an/[an+2^n]等式两边同时除以2^(n+1)a(n+1)/2^(n+1)=2^(n+1)an/[2^(n+1)(an+2^n)]a(n+1)/2^(n+1)=an/(an+2^n)]取倒数2^(n+1)/a(n+1)=(an+2^n)/an2^(n+1)/a(n+1)=2^n/an+12^(n+1)/a(n+1)-2^n/an=1所以数列{2^n/an}是以1为公差的等差数列2^n/an=2^1/a1+n-12^n/an=2+n-12^n/an=n+1取倒数an=2^n/(n+1)bn=n(n+1)an=2^n/(n+1)*n(n+1)=n*2^nsn=1*2^1+2*2^2+3*2^3+..+n*2^n2sn=1*2^2+2*2^3+3*2^4+.+(n-1)*2^n+n*2^(n+1)sn-2sn=2^1+2^2+2^3+..+2^n-n*2^(n+1)-sn=2*(1-2^n)/(1-2)-n*2^(n+1)-sn=2^(n+1)-2-n*2^(n+1)sn=n*2^(n+1)-2^(n+1)+2=(n-1)*2^(n+1)+2