KCC's Quizzes : AQQ234 - Sum of even integers

n = 1; f(n) = 2 = 1 + 1^2

n = 2; f(n) = 6 = 2 + 2^2

n = 3' f(n) = 12 = 3 + 3^2

n = 4; f(n) = 20 = 4 + 4^2

...

So general formula is f(n) = n + n^2

Sum with n = 5298 is therefore 28,074,102

n = 1; f(n) = 2 = 1 + 1^2

n = 2; f(n) = 6 = 2 + 2^2

n = 3' f(n) = 12 = 3 + 3^2

n = 4; f(n) = 20 = 4 + 4^2

...

So general formula is f(n) = n + n^2

Sum with n = 5298 is therefore 28,074,102

Oh wait, maybe I haven't quite got this right.  My n is the number in the sequence.  If the n in the question refers to the actual even integer, then I would have to divide this by 2 to get the position in the sequence.  In that case, the answer would be 7,019,850, as 5298 is the 2649th number in the sequence of even integers >0.  In this case I concur with Gaetano.

Good on time catch Martin! Yes, it is easy to confuse the number in the sequence and the even integer itself! Congrats!