Find the sum of the given series  (1x2) + (2x3) + (3x4)……… [(n-1)n]  

       1²  =  1

      2²  =  2 + 1.2

    3²  =  3  + 2.3

    4²  = 4 + 3 x 4

    n² = n + (n-1)n

Summing up all the equations, ( 1² + 2² + 3² + 4² ……. n²) =( 1 + 2 + 3 + 4 ……. n)+ [(1x2) + (2x3) + (3x4)……… [(n-1)n]

We know , ( 1² + 2² + 3² + 4² ……. n²) = n(n+1)(2n+1)/6

( 1 + 2 + 3 + 4 ……. n) = n(n+1)/2

Hence , [(1x2) + (2x3) + (3x4)……… [(n-1)n] = n(n+1)(2n+1)/6  - n(n+1)/2

On simplification it reduces to n(n²-1) /3