(1x2) 1 + (2x3) 1 + +(2013x2014) 1 / 2 =? How to calculate this?

=(1-1 / 2) + (1 / 2-1 / 3) + (1 / 3-1 / 4) +. + (1 / 2013-1 / 2014)
=1-1 / 2014
=2013 / 2014

2/1X2+2/2X3+2/3X4+…… +2 / 2007 x2008 equal to how much (need process) urgent!
Be in a hurry

2/1X2+2/2X3+2/3X4+…… +2/2007X2008 =2X[(2-1)/1X2+(3-2)2/2X3+(4-3)/3X4+…… +(2008-2007)/2007X2008] =2X(1-1/2+1/2-1/3+1/3-1/4+…… +1/2007-1/2008] =2x(1-1/2008)=2*2007/2008=2007/1004

Mathematical Olympiad questions 1x2 + 2x3 + 3x4 + &; &; &; + 98x99 =?
1^_ 2+2^_ 2+3^_ 2+•••+99^_ 2=?

an＝n(n+1)＝n²+n,
∴Sn＝(1²+2²+3²+… +n²)+(1+2+3+… +n)
＝n(n+1)(2n+1)/6+n(n+1)/2
＝n(n+1)(n+2)/3.
∴1×1+2×3+3×4+… +98×99
＝98×99×100÷3
＝485100.

1）1x2+2x3+3x4+...+100x101= 2）1x2+2x3+3x4+...n(n+1)= （3）1x2x3+2x3x4+3x4x5+...+n（n+1)（n+2）=
Gauss, a great mathematician, once studied such a problem when he was in school, 1 + 2 + 3 +. + 100 =? After research, the general conclusion of this problem is: 1 + 2 + 3 +... + n = 1 / 2n (n + 1), where n is a positive integer. Now let's study a similar problem 1x2 + 2x3 + 3x4 +... + n (n + 1) =? Observe the following three special equations: 1x2 = 1 / 3 (1x2x3-0x1x2) 2x3 = 1 / 3 (2x3x4-1x2x3) 3x4 = 1 / 3 (3x4x5-2x3x4) Add the two sides of these three equations, and you can get: 1x2 + 2x3 + 3x4 = 1 / 3x3x4x5 = 20. After reading this passage, please think and answer: (1) 1x2 + 2x3 + 3x4 +... + 100x101 = 2) 1x2 + 2x3 + 3x4 +... N (n + 1) = guess the following formula according to the above result: (3) 1x2x3 + 2x3x4 + 3x4x5 +... + n (n + 1) (n + 2)=

(1)1x2+2x3+3x4+...+100x101
=1/3*(100*101*102-99*100*101+99*100*101.-0*1*2)
=1/3*102*100*101
=343400
(2)1x2+2x3+3x4+...n(n+1)=
=1/3*[n*(n+1)*(n+2)-(n-1)*n*(n+1).-0*1*2]
=1/3n(n+1)(n+2)
(3)1x2x3+2x3x4+3x4x5+...+n(n+1)(n+2)
=1/4[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)+(n-1)n(n+1)(n+2).-0*1*2*3)]
=1/4n(n+1)(n+2)(n+3)

(1x2) 1 + (2x3) 1 + +(2013x2014) 1 / 2 =? How to calculate this?