f(k)=1-1/2+1/3-1/4+……1/2k-1-1/2k,则f(k+1)=f(k)+?

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f(k)=1-1/2+1/3-1/4+……1/2k-1-1/2k,则f(k+1)=f(k)+?

f(k)=1-1/2+1/3-1/4+……1/2k-1-1/2k,则f(k+1)=f(k)+?
f(k)=1-1/2+1/3-1/4+……1/2k-1-1/2k,则f(k+1)=f(k)+?

f(k)=1-1/2+1/3-1/4+……1/2k-1-1/2k,则f(k+1)=f(k)+?
注意:f(k)是2k个数之和,f(k+1)就是2k+2个数之和,所以f(k+1)=1-1/2+1/3-1/4+……(1/2(k+1)-1)-1/2(k+1)=f(k)+1/2(k+1)-1)-1/2(k+1)=f(k)+1/(2k+1)(2k+2)

f(k)=1-1/2+1/3-1/4+……1/(2k-1)-1/2k
f(k+1)=1-1/2+1/3-1/4+……1/(2k-1)-1/2k+1/(2k+1)-1/(2k+2)
故f(k+1)-f(k)=1/(2k+1)-1/(2k+2)
f(k+1)=f(k)+1/(2k+1)-1/(2k+2)

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