设f(n)=1/(n+1)+1/(n+2)+1/(n+3)+…+1/(n+2^n),则f(k+1)-f(k)=是2的n次方不是2n啊

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设f(n)=1/(n+1)+1/(n+2)+1/(n+3)+…+1/(n+2^n),则f(k+1)-f(k)=是2的n次方不是2n啊设f(n)=1/(n+1)+1/(n+2)+1/(n+3)+…+1/

设f(n)=1/(n+1)+1/(n+2)+1/(n+3)+…+1/(n+2^n),则f(k+1)-f(k)=是2的n次方不是2n啊
设f(n)=1/(n+1)+1/(n+2)+1/(n+3)+…+1/(n+2^n),则f(k+1)-f(k)=
是2的n次方不是2n啊

设f(n)=1/(n+1)+1/(n+2)+1/(n+3)+…+1/(n+2^n),则f(k+1)-f(k)=是2的n次方不是2n啊
f(n)=1/(n+1)+1/(n+2)+1/(n+3)+…+1/(n+2n),f(n+1)=1/(n+2)+1/(n+3)+…+1/(n+2n)+1/(n+2n+1)
f(n+1)-f(n)=1/(n+2n+1)-1/(n+1)

f(k+1)-f(k)=1/(k+2k+1)-1/(k+1)