大师作文网1²+2²/1×2+2²+3²/2×3+……+2001²+2002&#
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1²+2²/1×2+2²+3²/2×3+……+2001²+2002²/2001×2002
/>考察一般项：
[n²+(n+1)²]/[n(n+1)]
=(2n²+2n+1)/(n²+n)
=2 +1/(n²+n)
=2+1/[n(n+1)]
=2+1/n -1/(n+1)
(1²+2²)/(1×2)+(2²+3²)/(2×3)+……+(2001²+2002²)/(2001×2002)
=2+1/1-1/2+2+1/2-1/3+...+2+1/2001-1/2002
=2×2001+(1-1/2+1/2-1/3+...+1/2001-1/2002)
=4002+(1-1/2002)
=4002+2001/2002
=8014005/2002,也是4002又2002分之2001.
[n²+(n+1)²]/[n(n+1)]
=(2n²+2n+1)/(n²+n)
=2 +1/(n²+n)
=2+1/[n(n+1)]
=2+1/n -1/(n+1)
(1²+2²)/(1×2)+(2²+3²)/(2×3)+……+(2001²+2002²)/(2001×2002)
=2+1/1-1/2+2+1/2-1/3+...+2+1/2001-1/2002
=2×2001+(1-1/2+1/2-1/3+...+1/2001-1/2002)
=4002+(1-1/2002)
=4002+2001/2002
=8014005/2002,也是4002又2002分之2001.
求2003²-2002²+2001²-2000²+……+3²-2
1²+2²/1×2+2²+3²/2×3+……+2001²+2002&#
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