Question

数学高手 请举手之劳 数学题:设S1=1+1/1^2+1/2^2,S2=1+1/2^2+1/3^2,S3=1+1/3^2

Answer 1

因为Sn=1+1/n^2+1/(n+1)^2=(n^4+2n^3+3n^2+2n+1)/(n^2*(n+1)^2)=(n*(n+1)+1)^2/(n^2*(n+1)^2)

所以√Sn=√(n*(n+1)+1)^2/(n^2*(n+1)^2)=[n(n+1)+1]/[n(n+1)]

所以S=3/2+7/6+13/12+...+[n(n+1)+1]/[n(n+1)]=(1+1/2)+(1+1/6)+...+(1+1/(n(n+1)))=n+[1/2+1/6+...+1/(n(n+1))]=n+[(1-1/2)+(1/2-1/3)+...+(1/n-1/(n+1))]=n+1-1/(n+1)