<s>[INST] How to evaluate a binomial sum with $2n$ in the exponent. The question is to evaluate the sum  $$\sum_{k=0}^n\binom nk 3^{2n-k}$$
have tried fitting into the binomial form of $\binom nk \times x^k\times y^{n-k}$
but I can't seem to bring it to the correct form. 
 [/INST] Using @lulu's hint:
\begin{align}
\sum_{k = 0}^{n} \binom{n}{k} 3^{2n - k}
& = \sum_{k = 0}^{n} \binom{n}{k} 3^{n + n - k}
= 3^{n} \sum_{k = 0}^{n} \binom{n}{k} 3^{n - k} \\
& = 3^n \sum_{k = 0}^{n} \binom{n}{k} 1^k 3^{n - k}
= 3^n (1 + 3)^n
= 3^n 4^n
= 12^n.
\end{align}
Alternatively, one could pull out the $3^{2n}$:
\begin{align}
\sum_{k = 0}^{n} \binom{n}{k} 3^{2n - k}
& = 3^{2n} \sum_{k = 0}^{n} \binom{n}{k} 3^{- k} 
= 9^{n} \sum_{k = 0}^{n} \binom{n}{k} 1^{n - k} \left(\frac{1}{3}\right)^{k} \\
& = 9^n \left(1 + \frac{1}{3}\right)^n
= 9^n \frac{4^n}{3^n}
= 3^n 4^n
= 12^n.
\end{align}
</s>