1X2+2X3+3X4+4X5...+2011X2013=
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/28 08:14:40
1X2+2X3+3X4+4X5...+2011X2013=
1X2+2X3+3X4+4X5...+2011X2013=
1X2+2X3+3X4+4X5...+2011X2013=
n(n+1)=n²+n
1*2=1²+1
2*3=2²+2
3*4=3²+3
-----
n(n+1)=n²+n
1X2+2X3+3X4+4X5...+n(n+1)
=(1²+2²+3²+----+n²)+(1+2+3+---+n)
=n(n+1)(2n+1)/6+n(n+1)/2
=(1/6)n(n+1)(2n+1+3)=n(n+1)(n+2)/3
把n=2012代入得:
1X2+2X3+3X4+4X5...+2012X2013
=2012*2013*2014/3=2012*671*2014=2719004728
1X2+2X3+3X4+4X5...+2012X2013
=1/3X1X2X3+1/3[2X3X4-1X2X3]+1/3[3X4X5-2X3X4]+....+1/3[2012X2013X2014-2011X2012X2013]
=1/3X2012X2013X2014
=2719004728
1X2+2X3+3X4+4X5...+2011X2013=
=1/3 (1×2×(3-0)+2×3×(4-1)+3×4×(5-2)+.....+2011×2012×(2013-2010))
=1/3× (1×2×3-0×1×2+2×3×4-1×2×3+3×4×5-2×3×4+......+2011×2012×2013-2010×2011×2012)
=1/3 × (2011×2012×2013)
=2714954572
裂项消元法 结果-2012
原式=1/(1/1-1/2) 1/(1/2-1/3) 1/(1/3-1/4) 1/(1/4......-1/2012 1/(1/2012-1/2013)=1-1/1/2013=-2012