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