由a1+a2+a3=9得3a2=9所以a2=3;由a2+a4+a6=21得3a4=21所以a4=7又因?yàn)閍4=a2+2d=3+2d.所以d=2,顯然可得a1=1.an=1+2(n-1)=2n-1因此bn=2^n*an=n2^(n+1)-2^n寫(xiě)出前四項(xiàng)排成一列,左邊加左邊,右邊加右邊可以看出sn=-2+2^3+2*2^4+3*2^5+...+(n-1)2^n+n2^(n+1) 令Un=2^3+2*2^4+3*2^5+...+(n-1)*2^n+n*2^(n+1)注意每一項(xiàng)與前一項(xiàng)的關(guān)系可想到倆邊乘2來(lái)提高2的指數(shù),2Un=2^4+2*2^5+3*2^6+...+(n-1)*2^（n+1）+n2^(n+2),倆式相減Un=-2^3-2^4-2^5+...-2^（n+1）+n*2^(n+2）=2^3*[1-2^(n-1)]+n*2^(n+2）=8+(n-1)*2^(n+2)所以sn=-2+Un=6+(n-1)*2^(n+2)