已知：an=3n-1，bn=2^n，求数列{anbn}的前n项和

显示全部楼层 · 2009-2-27 10:01:23

anbn=(3n-1)*2^nSn=(3*1-1)*2^1+(3*2-1)*2^2+(3*3-1)*2^3+……+[3*(n-1)-1]*2^(n-1)+(3*n-1)*2^n则2Sn=(3*1-1)*2^1*2+(3*2-1)*2^2*2+(3*3-1)*2^3*2+……+[3*(n-1)-1]*2^(n-1)*2+(3*n-1)*2^n*2=(3*1-1)*2^2+(3*2-1)*2^3+(3*3-1)*2^4+……+[3*(n-1)-1]*2^n+(3*n-1)*2^(n+1)下式减去上式：Sn=-(3*1-1)*2^1-3*2^2-3*2^3-3*2^4-……-3*2^n+(3*n-1)*2^(n+1)=-4+(3*n-...

千问 · 2009-2-27 10:01:23

cn=anbn=(3n-1)*2^nSn=2*2^1+5*2^2+……+(3n-1)*2^n2Sn=
2*2^2+……+(3n-4)*2^n+(3n-1)*2^(n+1)相减：Sn=(3n-1)*2^(n+1)-3*(2^2+2^3+……+2^n)-2*2^1=(3n-1)*2^(n+1)-3*[2^2-2^(n+1)]/(1-2...