若f(n)=1/1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+n),则f（k+1）-f(k)= ,f(1)=是用数列归纳法的，答得好给悬赏金的hin52过程给下阿

偶也想问你！

f（k+1）-f(k)=1/(1+2+3+...+n+n+1)，f(1)=1

1+2+3+...+n=n(n+1)/2
所以：f(n)=1/1+2/(2*3)+2/(3*4)+...+2/n(n+1)
=2[1-1/2+1/2-1/3+1/3-1/4+...+1/n-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)
f(k+1)-f(k)=2(k+1)/(k+2)-2k/(k+1)=2/(k+1)(k+2)
f(1)=1