已知数列an}=2^n,若bn=-an*log2(an),Sn=b1+b2+...+bn,求使Sn+n*2^(n+1)>50成立的_数学解答

已知数列an}=2^n,若bn=-an*log2(an),Sn=b1+b2+...+bn,求使Sn+n*2^(n+1)>50成立的正整数n的最小值

人气：298 ℃　时间：2020-03-24 13:35:30

解答

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),
50 < S(n) + n*2^(n+1) = 2[2^n-1],
25 < 2^n - 1,
26 < 2^n,
n=5