Inverse of an exponential summatory It's the first time for me in this website, and I am never been that good in maths. Anyway, I am trying to figure out if it is possible to find a way to invert an exponential summatory. Saying I have a value $A_{tot}$ which is given by
$$A_{tot} = \sum_{i=1}^N (b_ic)^2$$
where $c$ is a constant, and $b_i$ an $i^{th}$ random integer number (unfortunately, I don't know if they are normally distributed or not, does it matter?).
I'd like to derive (even approximated) $A_{tot}$, given I know the value of the sum of the $b$
$$b_{tot} = \sum_{i=1}^N b_i$$
and that each $b_i$ is unknown.
So far I got stuck with resolving something like $\sqrt{b_1c}+ \sqrt{b_2c}+ ...+ \sqrt{b_Nc}$, which (as far as I know), is not possible to simplify by any property.
Is it that all, or is there still something I can do? Thanks in advance.

EDIT: The answer below refers to a previous edit of the question.
The sum you are after is $$A_{tot} = \sum_{i=1}^N (ic)^2 = (1c)^2 + (2c)^2+\ldots + (Nc)^2 \\=c^2(1^2+2^2+\ldots + N^2) = c^2\left(\frac{N(N+1)(2N+1)}{6}\right)$$
since the sum of squares is given by $\sum_{i=1}^N i^2 = \frac{N(N+1)(2N+1)}{6}$.
If you are looking for an inverse, i.e. if given $A_{tot}$ what is $N$. Then we can use that for large $N$ then $N+1 \approx N$ and $2N+1\approx 2N$ so $$A_{tot} \approx c^2 \frac{N^3}{3}$$ which gives us the formula
$$N \approx \left\lfloor\left(\frac{3A_{tot}}{c^2}\right)^{1/3}\right\rfloor$$
where $ \lfloor x\rfloor$ is the floor-function (it removes the decimals of $x$ and leaves an integer value, e.g $ \lfloor 3.7\rfloor = 3$ and $ \lfloor 19.2\rfloor = 19$). It turns out that this formula is exact for all $N\geq 1$ (as long as $A_{tot}$ is the result of such a summation).
For example with $c=1$ and $N=3$ we have $A_{tot} = 1 + 2^2 + 3^2 = 14$. Now $\left(\frac{3A_{tot}}{c^2}\right)^{1/3} = 3.48$ and $\lfloor 3.48\rfloor = 3 = N$ showing that it works.
