$\sum_{k=0}^n\binom nk x^k=\sum_{k=0}^n\binom nk x^{n-k}.$ $$\sum_{k=0}^n\binom nk  x^k=\sum_{k=0}^n\binom nk  x^{n-k}.$$
I want a deeper understanding of solving problems in this nature. I can´t grasp this writing way.I get confused just by looking at these problems. Please be patient with me. ;)
 A: Hint
$${n\choose k}={n\choose n-k}$$
A: The sum on right hand side (RHS) of the equation is the sum on the left hand side (LHS) of the equation written backwards. Let me explain:
By definition of the sum,
$$
\sum_{k=0}^n\binom nk  x^k = \binom n0 x^0 + \binom n1 x^1 \ \ + \ \ ... \ \ +   \binom {n}{n-1} x^{n-1}\ + \binom {n}{n} x^{n} \\
$$
writing the sum in reverse order to yeild,
$$
= \binom {n}{n} x^{n} + \binom {n}{n-1} x^{n-1} \ \ + \ \ ... \ \ + \binom n1 x^1 +\binom n0 x^0\\
$$
with a bit more work we have,
$$
= \binom {n}{n-0} x^{n-0} + \binom {n}{n-1} x^{n-1}\ \ + \ \ ... \ \ + \binom n{n-(n-1)} x^{n-(n-1)} +\binom n{n-n} x^{n-n}\\
$$
that is,
$$
\sum_{k=0}^n\binom nk  x^k=\sum_{k=0}^n\binom n{n-k}  x^{n-k}.
$$
Finally we note the following combinatorial identity,
$$
\begin{align}
& \binom nk = \frac{n!}{k!(n-k)!} = \frac{n!}{(n-(n-k))!(n-k)!} = \binom n{n-k}
\end{align}
$$
Applying the combinatorial identity delivers the desired result,
$$
\sum_{k=0}^n\binom nk  x^k=\sum_{k=0}^n\binom nk  x^{n-k}.
$$
