最好的方法在这里!
当x=0时,yn=1/n;
当y=0时,xn=1/(n+1)
∴Sn=(1/2)·(1/n)·[1/(n+1)]=1/[2n(n+1)]
⑴当n=1时,代入前面结论得S1=1/4
⑵S1+S2+…+S6=1/(2·1·2)+1/(2·2·3)+1/(2·3·4)+…+1/(2·6·7)
==(1/2)·[1/(1·2)+1/(2·3)+1/(3·4)+…+1/(6·7)]
=(1/2)·[(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+…+(1/6-1/7)]
=(1/2)·(1/1-1/7)
=3/7