设n∈正整数,求证:1/2²+1/4²+…+1/(2n)²<1
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设n∈正整数,求证:1/2²+1/4²+…+1/(2n)²<1
设n∈正整数,求证:1/2²+1/4²+…+1/(2n)²<1
设n∈正整数,求证:1/2²+1/4²+…+1/(2n)²<1
当n=1时,原和式=1/2^2=1/4<1;当n≥2时,原和式=1/2^2+1/4^2+...+1/(2n)^2=(1/4)[1/1^2+1/2^2+...+1/n^2]<(1/4)[1/1+1/(2×1)+1/(3×2)+...+1/n(n-1)]=(1/4)[1+1/1-1/2+1/2-1/3+...+1/(n-1)-1/n]=(1/4)[2-1/n]<1/2<1;所以对n∈正整数,原和式成立.