一道分数裂项题，跪求答案和过程！！

因为1^2+2^2=(1+2)^2-2*1*2
2^2+3^2=(2+3)^2*2*2*3
.........
99^2+100^2=(99+100)^2-2*99*100
所以原式=[（1+2)^2-2*1*2]/1*2+[(2+3)^2-2*2*3]/2*3+....+[(99+100)^2-2*99*100]/99*100
=(3^2-2*1*2)/1*2+(5^2-2*2*3)/2*3+...+(199^2-2*99*100)/99*100
=3^2/1*2+5^2/2*3+7^2/3*4+...+199^2/99*100-2*99
=3^2(1/1-1/2)+5^2(1/2-1/3)+7^2(1/3-1/4)+....+199^2(1/99-1/100)-198
=3^2/1-3^2/2+5^2/2-5^2/3+7^2/3-7^2/4+...+199^2/99-199^2/100-198
=3^2+(5^2-3^2)/2+(7^2-5^2)/3+(9^2-7^2)/4+....+(199^2-197^2)/99-199^2/100-198
=9+(5+3)(5-3)/2+(7+5)(7-5)/3+(9+7)(9-7)/4+...+(100+197)(199-197)/99-199^2/100-198
=9+8+8+8+...+8-199^2/100-198
=9+98*8-199^2/100-198
=595-396.01
=198.99