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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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  • $\begingroup$ How did you come up with the value x=-3? $\endgroup$
    – Aaron
    Dec 14 '12 at 4:58
  • $\begingroup$ @Aaron, JohnD guessed it.The whole thing is pretty obvious once you recognize a pattern-a binomial expansion. $\endgroup$ Dec 14 '12 at 5:03
  • $\begingroup$ Look at the terms in your sum: the binomial coefficients are being multiplied by $(-3)^k$, $k=0,1,2,\dots,n$ each time. $\endgroup$
    – JohnD
    Dec 14 '12 at 5:04
  • $\begingroup$ Can you explain the alternating postive and negative sign? Or is it just for the second term $\endgroup$
    – Aaron
    Dec 14 '12 at 5:12
  • $\begingroup$ @Aaron: $(-3)^n=(-1)^n\cdot 3^n$, that's why you are seeing the alternating signs and increasing powers of 3. $\endgroup$
    – JohnD
    Dec 14 '12 at 16:21
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Recall that $$(1-x)^n = \sum_{k=0}^n \dbinom{n}k (-x)^k$$

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