Question

1*3/1+3*5/1+5*7/1+···+2013*2015/1=？

1*3/1+3*5/1+5*7/1+7*9/1+···+2013*2015/1=？

Answer 1

答：
利用分数的裂项知识求解
1/(1×3)+1/(3×5)+1/(5×7)+....+1/(2013×2015)
=(1/2)× ( 1-1/3+1/3-1/5+1/5-1/7+.,,,,,,+1/2013-1/2015)
=(1/2)×(1-1/2015) （中间各项正负互相抵消为0）
=(1/2)×(2014/2015)
=1007 / 2015
=2015分之1007

Answer 2

1/(1*3)+1/(3*5)+1/(5*7)+……+1/(2013*2015)
=1/2*(1-1/3)+1/2*(1/3-1/5)+1/2*(1/5-1/7)+……+1/2*(1/2013-1/2015)
=1/2*(1-1/3+1/3-1/5+1/5-1/7+……+1/2013-1/2015)
=1/2*(1-1/2015) 【中间的都约掉了】
=1/2*2014/2015
=1007/2015

Answer 3

1/1*3+1/3*5+1/5*7+...+1/2013*2015
=(1-1/3)/2+(1/3-1/5)/2+(1/5-1/7)/2+...+(1/2013-1/2015)/2
=[(1-1/3)+(1/3-1/5)+(1/5-1/7)+...+(1/2013-1/2015)]/2
=(1-1/2015)/2
=1007/2015

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Answer 4

原式=（2^2-1）+（4^2-1）+（6^2-1）+......+(2014^2-1)
=2^2(1²＋2²＋3²＋……＋1007^2)-1007
=4x1007x1008x2015÷6-1007
=1363557553

（1²＋2²＋3²＋……＋n²=n(n+1)(2n+1)/6）