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With $n$ a positive integer, evaluate the sum

$\binom{n}{0}-3\binom{n}{1}+3^2\binom{n}{2}+\cdots+(-1)^n3^n\displaystyle\binom{n}{n}=\sum_{k=0}^n(-3)^k\binom{n}{k}$

Anyone know how to approach this problem?

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From the definition of binomial coefficients, $\sum_{k=0}^n {n\choose k}x^k=(1+x)^n.$ For your problem, take $x=-3$ to conclude the sum is $(-2)^n$.

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    @Aaron: $(-3)^n=(-1)^n\cdot 3^n$, that's why you are seeing the alternating signs and increasing powers of 3.2012-12-14
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Recall that $(1-x)^n = \sum_{k=0}^n \dbinom{n}k (-x)^k$