an=(2^n)*n,求和Sn
问题描述:
an=(2^n)*n,求和Sn
答
sn=2+2²×2+2三次方×3+.+(2^n)*n……①
所以
2sn= 2²+.+(2^(n+1))n……②
①-②=-(2^(n+1))n+2+2²+……+2n次方
-sn=-(2^(n+1))n+2(1-2n次方)/(1-2)
sn=2(n+1)次方(n-1)+2