Calculate Sum[ k(k+3)] for k = 1 to k = n.

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We have to find the sum of k(k+3) with k = 1 to k = n.

Now Sum[ k(k+3)]

=> Sum [ k^2 + 3k]

=> Sum [k^2] + 3*Sum [k]

For k equal to 1 to n

=> n(n+1)(2n+1)/6 + 3*n*(n+1)/2

=> n(n+1)(2n+1)/6 + (3/2)*n*(n+1)

Therefore the required sum is n(n+1)(2n+1)/6 + (3/2)*n*(n+1).

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