s=1/2+1/(2+4)+1/(2+4+6)+1/(2+4+6+8)+...+1/(2+4+6+8+...+2n)=?

s=1/2+1/(2+4)+1/(2+4+6)+1/(2+4+6+8)+...+1/(2+4+6+8+...+2n)=?
s=1/2+1/(2+4)+1/(2+4+6)+1/(2+4+6+8)+...+1/(2+4+6+8+...+2n)=?

s=1/2+1/(2+4)+1/(2+4+6)+1/(2+4+6+8)+...+1/(2+4+6+8+...+2n)=?
答案:n/(n+1)
解:(用裂项相消法)
s=1/2+1/(2+4)+1/(2+4+6)+1/(2+4+6+8)+...+1/(2+4+6+8+...+2n)
=1/2+1/6+1/12+1/20+…+1/(2+4+6+8+…+2n)
=1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+…+1/[n*(n+1)]
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+…+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)