s=1/2+1/(2+4)+1/(2+4+6)+1/(2+4+6+8)+...+1/(2+4+6+8+...+2n)=?
问题描述:
s=1/2+1/(2+4)+1/(2+4+6)+1/(2+4+6+8)+...+1/(2+4+6+8+...+2n)=?
答
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-...