(1²/2002-2²/2002+.13²/2002-14²/2002+15²/2002

(1²/2002-2²/2002+.13²/2002-14²/2002+15²/2002
(1²/2002-2²/2002+.13²/2002-14²/2002+15²/2002

(1²/2002-2²/2002+.13²/2002-14²/2002+15²/2002
(1²/2002-2²/2002+.13²/2002-14²/2002+15²/2002
=（1²-2²+3²-4²+.+13²-14²+15²）/2002
=(1+(3-2)(3+2)+(5-4)(5+4)+.+(15-14)(15+14))/2002
=(1+2+3+...+14+15)/2002
=【(1+15)×15÷2】/2002
=120/2002
=60/1001

=[1+(3²-2²)+(5²-4²)+……+(15²-14²)]/2002
=[1+(3+2)(3-2)+……+(15+14)(15-14)]/2002
=(1+2+3+……+15)/2002
=(15×16÷2)/2002
=60/1001

1²/2002-2²/2002+......13²/2002-14²/2002+15²/2002
=(1²-2²+3²-4²+...+13²-14²+15²)/2002
=[1+(3²-2²)+(5²-4²)+...+(15²-14²)]/2002
=(1+2+3+...+14+15)/2002
=120/2002
=60/1001

反括号呢……

显然分母都为2002，现在来讨论分子
奇数项和为1^2+3^2+5^2+7^2+9^2+11^2+13^2+15^2=680
偶数项和为-(2^2+4^2+6^2+8^2+10^2+12^2+14^2)=560
所以奇数项和-偶数项和=680-560=120
所以(1²/2002-2²/2002+......13²/2002-14²/2002+15²/2002=120/2002=60/1001