Question

已知an=2^n,bn=an×log½an,求数列bn的前n项和。

Answer 1

an=2^n
∴bn=an×log?an=2^n×log?(2^n)
=2^n×nlog? (2)
=-n*2^n
∴Sn=-[1*2^1+2*2^2+3*2^3……+n*2^n] ①
∴ 2Sn= -[1*2^2+2*2^3+3*2^4……+n*2^(n+1)] ②
∴②-①得Sn=(2^1+2^2+2^3+……+2^n)-n*2^(n+1)
=2^(n+1)-2-n*2^(n+1)
=(1-n)2^(n+1) -2

Answer 2

b(n)=-n2^n
b(n+1)-b(n)=-(n+1)2^(n+1)+n2^n=-n2^n-2^n=-b(n)-2^n
……
b(2)-b(1) = -b(1)-2^1
两边求和
b(n+1)-b(1) = -S(n)-2(1-2^n)/(1-2)
(n+1)2^(n+1)-2=-S(n)-2^(n+1)+2
S(n) = -(n+2)·2^(n+1) + 2