另x=0,y= 1÷(k+1);y=0,x= -1÷k
故S=1/2÷(k+1) ÷k
=1/2*(1/k-1/(k-1))
显然分母为等差数列,只要错位求和,很容易得到:
s1+s2+``````sn=1/2(1/(1*2)+1/(2*3)+``````1/(n(n+1)))
=1/2(1-1/2+1/2-1/3+``````1/n-1/(n+1))
=1/2(1-1/(n+1))
=n/2(n+1)
另x=0,y= 1÷(k+1);y=0,x= -1÷k
故S=1/2÷(k+1) ÷k
=1/2*(1/k-1/(k-1))
显然分母为等差数列,只要错位求和,很容易得到:
s1+s2+``````sn=1/2(1/(1*2)+1/(2*3)+``````1/(n(n+1)))
=1/2(1-1/2+1/2-1/3+``````1/n-1/(n+1))
=1/2(1-1/(n+1))
=n/2(n+1)