Vikas TU
Last Activity: 8 Years ago
1*2+2*3+3*4+...............n
ca be done by summation series.
Observe the pattern carefully in terms of n starting with n=1.
=> summation{from n=1 to n} = n(n+1) = summation(n^2 + n)
=> sigma(n^2) + sigma(n)
=> n(n+1)(2n+1)/6 + n(n+1)/2
=> n(n+1){(2n+1)/3+ 1}
=> n(n+1)(2n + 4)/3