Question

已知数列{an}=2^n,若bn=an*log1/2(an),Sn=b1+b2+...+bn,求使sn+n*2^(n+1

Answer 1

a(n)=2^n,

b(n)=-a(n)ln[a(n)]/ln2=-2^nln[2^n]/ln2=-n2^n,

S(n) = b(1)+b(2)+...+b(n)=-1*2 -2*2^2 - 3*2^3 - ...- (n-1)*2^(n-1) - n*2^n,

2S(n) = -1*2^2 - 2*2^3 - 3*2^4 -...- (n-1)*2^n - n*2^(n+1),

S(n)=2S(n)-S(n)=2 + 2^2 + 2^3 + ...+ 2^n - n*2^(n+1)

=2[2^n-1]/(2-1) - n*2^(n+1),

= 2[2^n-1] - n*2^(n+1),

S(n) + n*2^(n+1) = 2[2^n-1]=30

n=4