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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.

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    $\begingroup$ Hint: The $2n$ in the exponent is irrelevant. just pull $3^{2n}$ out of the sum. $\endgroup$
    – lulu
    Mar 4, 2019 at 0:21

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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}

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    $\begingroup$ Or even simpler $3^{2n-k}=9^{n-k}3^k\to (9+3)^n=12^n$. $\endgroup$
    – zwim
    Mar 4, 2019 at 0:39

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