Sum of a series

The numbers repeat themselves after the $\ldots + n + \ldots$ part so all you have to do is find twice the sum of

$1 + 2 + 3 + \ldots + (n - 1)$

and then just add $n$

Speaking of sums, the formula for working out the sum of a series is

$\displaystyle \sum_{1}^{x} = \frac {x(x + 1)}{2}$

So by slotting that into the series you get

$2\frac {(n - 1)(n - 1 + 1)}{2} + n$

See now you have something to work with, the next step is to simplify and expand

$(n - 1)(n - 1 + 1) + n \Rightarrow n(n - 1) + n \Rightarrow n^2 - n + n$

Can you see where this is going

$n^2 = 419,904$

So $n = \sqrt {419,904} = 648$

Next is the prime factors of $648$

$648 = 324 \cdot 2 = 162 \cdot 2 \cdot 2 = 81 \cdot 2 \cdot 2 \cdot 2$

We can't divide by $2$ any more but we can divide by $3$

$81 = 27 \cdot 3 = 9 \cdot 3 \cdot 3 = 3 \cdot 3 \cdot 3 \cdot 3$

This means that the prime factors of $648$ are $2, 2, 2, 3, 3, 3, 3$

Adding these together gives you

$(2 \cdot 3) + (3 \cdot 4) = 6 + 12 = 18$

So the answer is $18$