求解几个初一数学题

显示全部楼层 · 2009-10-5 17:36:17

11/[n*(n+1)]2 规律同上，因1/[n*(n+1)]=1/n-1/(n+1)所以=1/1-1/2 +1/2-1/3+...+1/n-1/(n+1)
=1-1/(n+1)由n*(n+1)=4018020求得n = 2004所以结果为=1-1/2005=2004/20053 =1/2-(...)=1/2-(1/2-1/6-1/40) = 1/6+1/40 = (20+3)/120=23/120

千问 · 2009-10-5 17:36:17

N分之1×（N+1）分之1

千问 · 2009-10-5 17:36:17

11/[n*(n+1)]2 因为1/[n*(n+1)]=1/n-1/（n+1）所以 1/2+1/6+1/12+）……+（1/4018020）
=1/1-1/2 +1/2-1/3+...+1/n-1/(n+1)
=1-1/(n+1) 由n*(n+1)=4018020 求得n = 2004所以结果为=1-1/2005=2004/20053利用上面的规律
1/2-1/6-1/12-1/20-1/30-1/40
=1/2-(1/2-1/3)-(1/3-1/4)-(1/4-1/5)-(1/5-1/6)-1/40
=1/2-1/2+1/3-1/3+1/4-1/4+1/5-1/5+1/6-1/40
=1/6-1/40
=17/120