ײ+2×+1=4一元二次方程

1个回答

  • 令an=n(n+1)=n²+n

    ∴1×2+2×3+3×4+...+2011×2012

    =1²+1+2²+2+3²+3+……+2011²+2011

    =﹙1²+2²+3²+……+2011²﹚+﹙1+2+3+……+2011﹚

    =2011*2012*4023÷6+2011*2012/2

    1/(2011x2012)x(1x2+2x3+3x4+……+2011x2012)

    =1/(2011x2012)x[(1²+2²+3²+……+2011²)+(1+2+3+……+2011)]

    =1/2011x2012x[2011x(2011+1)(2x2011+1)/6+2011x(1+2011)/2]

    =1/2012x(2012x4023/6+2012/2)

    =4023/6+1/2

    =1341/2+1/2

    =671

    另一种方法:很明显,这里面最复杂的就是1×2+2×3+3×4+……+2011×2012

    首先告诉你,1×2+2×3+3×4+……+n(n+1)=1/3×n(n+1)(n+2)

    故括号内为1/3×2011×2012×2013

    与前面的相乘就是1/3×2013=671

    满意请采纳.