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

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!

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