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设f(n)=1/(n+1)+1/(n+2)+……+1/(2n) (n∈N新),那么f(n+1)-f(n)等于(1/(2n

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设f(n)=1/(n+1)+1/(n+2)+……+1/(2n) (n∈N新),那么f(n+1)-f(n)等于(1/(2n+1) - 1/(2n+2))
设f(n)=1/(n+1)+1/(n+2)+……+1/(2n) (n∈N新),那么f(n+1)-f(n)等于(1/(2n
f(n)=1/(n+1)+1/(n+2)+……+1/(2n)
f(n+1)= 1/(n+2)+……+1/(2n)+1/(2n+1)+1/(2n+2)
f(n+1)-f(n)=1/(2n+1)+1/(2n+2)-1/(n+1)
=1/(2n+1)+1/(2n+2)-2/(2n+2)
=1/(2n+1)-1/(2n+2)