I have a summation series as following: $$\sum_{n=1}^N f(n)=-\sum_{n=1}^Ng(n)+\frac{X}{4}-\sum_{i=1}^3\bigg|\sum_{n=1}^N h_i(n)\bigg|^2$$ where $N$ is some known integer and $X$ is a constant.
Question:
Is there any way by which I can write $f(n)$ in terms of $g(n)$ and $h_i(n)$ i.e. I want to get rid of summation sign on LHS of equation.
My attempt:
Write last term as: $$\bigg|\sum_{n=1}^N h_i(n)\bigg|^2=\sum_{n=1}^N h^*_i(n) \sum_{n=1}^N h_i(n)$$ so $$\sum_{n=1}^N f(n)=-\sum_{n=1}^Ng(n)+\frac{X}{4}-\sum_{i=1}^3\bigg\{\sum_{n=1}^N h^*_i(n) \sum_{n=1}^N h_i(n) \bigg\}\\ \sum_{n=1}^N f(n)=\sum_{n=1}^N \bigg[-g(n)+\frac{X}{4}-\sum_{i=1}^3 \bigg\{h^*_i(n) \sum_{n=1}^N h_i(n)\bigg\}\bigg]\\ f(n)=\bigg[-g(n)+\frac{X}{4}-\sum_{i=1}^3 \bigg\{h^*_i(n) \sum_{n=1}^N h_i(n)\bigg\}\bigg]$$
I am not sure if I am doing it right or not. Also, $\frac{X}{4}$ factor was not depending on $n$.
No, there isn't: You have an equation constraining the sum $f(1) + f(2) + \cdots + f(N)$ to some value that depends on the functions and constant on the RHS. All you have is a constraint on the sum, which is not sufficient to obtain equations for the individual elements $f(i)$ (except in the trivial case where $N=1$). A simple analogy to this would be if I told you that I have a total of $245 in my money box, and asked you to deduce the individual notes/coins that make up that sum (with no additional information other than the number of notes + coins). Obviously that information would be insufficient to answer my question - you cannot deduce individual values solely from a constraint on their sum.
Note: For the above, I am taking you at your word that this equation holds true for some particular value of $N$ (i.e., that it is a single constraint). If, on the other hand, you mean that this equation holds for all $N \in \mathbb{N}$ then that is actually a whole sequence of equations, and in that case you could get equations for each $f(i)$ via induction.