Dealing with non-constant term in Binomial Theorem question

Question

I am wondering this. Suppose I have a sequence $\{\varepsilon_n\}_{n=0}^\infty$ and elements of this sequence are part of a binomial type expression: For example, my expression is $$\sum_{k=0}^n\binom{n}{k}\varepsilon_k2^{n-k}$$

If I knew, for example that the expresssion was equal to 1, then I would know that $\varepsilon_n=(-1)^n$ because

$$\sum_{k=0}^n\binom{n}{k}\varepsilon_k2^{n-k}=1=1^n=[2+(-1)]^n=\sum_{k=0}^n\binom{n}{k}(-1)^k2^{n-k}$$

I'm almost sure that the answer is "no" but here it is. Is there any way to simplify an expression like mine as

$$(2+x)^n=\sum_{k=0}^n\binom{n}{k}\varepsilon_k2^{n-k}$$ when the terms in the expansion are not constants such as $x$ raised to powers but corresponding elements indexed appropriately?

Answer 1

Please note, that the following statement is not valid.

If I knew, for example that the expresssion was equal to $1$, then I would know that $\varepsilon_n=(-1)^n$ because \begin{align*} \sum_{k=0}^n\binom{n}{k}\varepsilon_k2^{n-k}=1=1^n=[2+(-1)]^n=\sum_{k=0}^n\binom{n}{k}(-1)^k2^{n-k}\tag{1} \end{align*}

The knowledge that $(-1)^k$ with $0\leq k \leq n$ in the RHS of (1) give \begin{align*} \sum_{k=0}^n\binom{n}{k}(-1)^k2^{n-k}=1\tag{2} \end{align*} is not sufficient to conclude that \begin{align*} \sum_{k=0}^n\binom{n}{k}\varepsilon_k2^{n-k}=1\tag{3} \end{align*} already implies $\varepsilon_k=(-1)^k$ for all $k$.

In fact the LHS of 3 has $n+1$ free parameter $\varepsilon_k$. So, you could put $\varepsilon_0$ to $\varepsilon_{n-1}$ to whatever you like they should be. Than you can take $$\varepsilon_n=1-\sum_{k=0}^{n-1}\binom{n}{k}\varepsilon_k2^{n-k}$$ and the equation (1) is fulfilled.

Conclusion: A simplification of the sum (3) in the general case (arbitrary $\varepsilon_k, 0\leq k \leq n-1$) is not achievable.