Ideal Reconstruction of Upsampled Signal Problem: The signal $cos(2\pi14100t)$ is sampled at $F_s = 400 Hz$. It is then upsampled with a factor 3 and then reconstructed ideally with a new frequency $F = 500 Hz$.

I now want to find the new signal that is created.

Incomplete solution: With the frequency before sampling $\pm 141000 Hz$. After sampling the normalized frequencies:

$f = \pm {14100 \above 1 pt 400} \pm k = \pm (35 + {1 \above 1 pt 4})\pm k= \pm {1 \above 1 pt 4} \pm k $ per sample

Interpolation with factor 3 yields the normalized frequencies:

$f = {\pm {1 \above 1 pt 4} \pm k \above 1 pt 3} = $ 3 frequencies $\pm k?$

I know how to solve it past that to get the final $F= \pm f * F_s$. I.e. $y(t) = cos(2\pi f_1t) + cos(2\pi f_2t) + cos(2\pi f_3t)$.

So my question is how do I calculate the 3 frequencies? Would greatly appreciate a simple solution and not just an answer so that I actually understand this.
There are some mistakes in the question and your description.

When we sample a signal of frequency $f_m$ with a sampling rate $f_s$, the sampled signal contains the frequencies $f=f_m \pm nf_s$, where $n \in \mathbb{Z} $.
Here the frequencies available in the sampled signal are calculated as,

$f_m=14100 Hz$
 and $f_s=400 Hz$.
we know that, 

$ f=f_m \pm nf_s $, where $n = 0, \pm 1, \pm 2, \pm 3, ... $

ie., $ f= ..., -1500, -1100, -700, -300, 100, 500, 900, 1300, 1700, ... $

When we up sample the signal by 3, the frequencies will become $f/3$.

So here the new frequencies are, $ f_u = ..., -633.33, -500, -366.67, -233.33, -100, 33.33, 166.67, 300, 433.33, 566.67, ... $

After upsampling the signal is reconstructed by using an LPF of cutoff frequency 500.
When we apply LPF to the above signal only the frequencies whose absolute value is below 500 remains and other frequencies will loss.

there for after applying LPF the frequency components available in the signal are,  $ -500, -366.67, -233.33, -100, 33.33, 166.67, 300, 433.33 $.

I hope this solves your problem.