Question

an=n?2^n,求an的前n项和

Answer 1

an=n?2^n 问号是加还是乘
如果是加Sn=n(n+1)/2+2(2^n -1)
如果是乘
Tn=1x2^1 + 2x2^2 + 3x2^3 + ...+ (n-1)x2^(n-1) + n x2^n
2Tn= 1x2^2 + 2x2^3 + ...+ (n-2)x2^(n-1) + (n-1)x2^n + nx2^(n+1)
第二个式子减第一个式子（错位相减法）
Tn = 2Tn - Tn
=-1x2^1 - 1x2^2 - 1x2^3 - ...- ( 1)x2^(n-1) - ( 1)x2^n + nx2^(n+1)
=nx2^(n+1)-[1x2^1+1x2^2 + 1x2^3 + ...+ 1x2^(n-1) + 1x2^n]
中括号里是一个等比数列,项数为n
=nx2^(n+1)-2x(2^n -1)
=nx2^(n+1)-2^(n+1)+2
=(n-2)x2^(n+1)+2