显然可以得到1/(1^2-2^2)=1/[(1+2)*(1-2)]=1/2 *(1/3 -1)1/(1^2-3^2)=1/[(1+3)*(1-3)]=1/2 *(1/4 -1/2)1/(1^2-4^2)=1/[(1+4)*(1-3)]=1/2 *(1/5 -1/3)……1/(1^2-100^2)=1/[(1+100)*(1-100)]=1/2*(1/101-1/99)于是相加抵消之后得到原式=1/2 *(-1-1/2+1/100+1/101)= -14949/10100