1x2/3+2x3/3+3x4/3...+2003x2004/3

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1x2/3+2x3/3+3x4/3...+2003x2004/3
1x2/3+2x3/3+3x4/3...+2003x2004/3

1x2/3+2x3/3+3x4/3...+2003x2004/3
首先说明一下,当分式左右书写的时候,分子写在“/”左边,分母写在“/”右边.
如果本题写的是对的,3是分母：
考察一般项：
n(n+1)/3=(1/3)(n²+n)
1×2/3+2×3/3+...+2003×2004/3
=(1/3)[(1²+2²+...+2003²)+(1+2+...+2003)]
=(1/3)(2003×2004×4007/6 +2003×2004/2)
=(1/3)(2680691014+2007006)
=2682698020/3
用到的公式：
1²+2²+...+n²=n(n+1)(2n+1)/6
1+2+...+n=n(n+1)/2
对于本题,n=2003.
如果是分子分母写反了的话,3是分子：
3/(1×2)+3/(2×3)+...+3/(2003×2004)
=3[1/(1×2)+1/(2×3)+...+1/(2003×2004)]
=3(1-1/2+1/2-1/3+...+1/2003-1/2004)
=3(1-1/2004)
=2003/668

1×2在分母...
3/(1×2)+3/(2×3)+...+3/(2003×2004)
=3*(1-1/2+1/2-1/3+...+1/2003-1/2004)
=3*(1-1/2004)
=3*2003/2004
=2003/668

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