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$$\sum_{k=0}^n {n \choose k} \cdot (-1)^k \cdot 3^{n-k}$$

I tried generating the series. However, I don't know what to do beyond that.

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    HINT: your sum is given by $$2^n\cdot 3^{n-1}$$ – 2017-01-05
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    @Dr.SonnhardGraubner Huh? – 2017-01-05

1 Answers 1

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Recall the binomial equality $$ \sum_{k=0}^n \binom nk x^k y^{n-k} = (x+y)^n $$ Now apply this for $x = -1$ and $y = 3$.