bn=1/(2n+1),b1平方+b2平方+...bn平方
问题描述:
bn=1/(2n+1),b1平方+b2平方+...bn平方
数学人气:804 ℃时间:2020-10-02 05:42:12
优质解答
bn=1/(2n+1)
bn^2cn=1/[2n(2n+1)]=1/2n-1/(2n+1)
b1^2+b2^2+...+bn^2n=2时,b1^2+b2^2n>2时,1/2n11/(n+1)>1/2b1^2+b2^2+...+bn^22n^2+2n1/(4n^2+4n+1)
bn^2cn=1/[2n(2n+1)]=1/2n-1/(2n+1)
b1^2+b2^2+...+bn^2n=2时,b1^2+b2^2n>2时,1/2n11/(n+1)>1/2b1^2+b2^2+...+bn^22n^2+2n1/(4n^2+4n+1)
答
bn=1/(2n+1)
bn^2cn=1/[2n(2n+1)]=1/2n-1/(2n+1)
b1^2+b2^2+...+bn^2n=2时,b1^2+b2^2n>2时,1/2n11/(n+1)>1/2b1^2+b2^2+...+bn^22n^2+2n1/(4n^2+4n+1)