Dealing with non-constant term in Binomial Theorem 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?  
 A: 
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.
