求数列{n×2^n}的前n项和sn

显示全部楼层 · 2013-4-27 21:37:45

sn=1*2^1+2*2^2+3*2^3+..........+n*2^n2sn=1*2^2+2*2^3+3*2^4+..........+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=n*2^(n+1)-2*(1-2^n)/(1-2)sn=n*2^(n+1)-2*(2^n-1)sn=n*2^(n+1)-2^(n+1)+2sn=(n-1)*2^(n+1)+2...

千问 · 2013-4-27 21:37:45

2Sn=1×2^2+2×2^3+3×2^4+....+n×2^(n+1)Sn=1×2+ 2×2^2+3×2^3+....+n×2^n2Sn-Sn=2+2×2+3×2+....+n×2
Sn=2(1+2+3+.....+n)
Sn=2×{(1+n)n/2}=(1+n)n希望能帮助你
望采纳为满意...

千问 · 2013-4-27 21:37:45

Sn=1×2^1+2×2^2+3×2^3+……+n×2^n……①2Sn=1×2^2+2×2^3+3×2^4+……+(n-1)×2^n+n×2^(n+1)……②②-①得 Sn=-1×2^1+(-2^2)+(-2^3)+……+(-2^n)+n×2^(n+1)=-(2^1+2^2+……+2^n)+n×2^(n+!)=-(2-2^(n+1))/(1-2)+n×...