Question

一道奥数题，高手解

[2/1*（1+2）]+[3/(1+2)(1+2+3)]+……[2000/（1+2+3+4+……+1999)(1+2+3+4+5+……+2000)]

Answer 1

即求通项公式为:an=(n+1)/(1+..+n)*(1+..+(n+1))的前1999项和
an=4(n+1)/n(n+1)(n+1)(n+2)
=4/n(n+1)(n+2)
=2[1/n(n+1) -1/(n+1)(n+2)]
所以前1999项和为:
Sn=2*[1/1*2 -1/2*3 +1/2*3 -1/3*4 +...+1/1999*2000 -1/2000*2001]
=2*(1/1*2-1/2000*2001)
=2*(1/2-1/2000+1/2001)
=1-2*(1/2000-1/2001)
=...
自己去化简吧