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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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up vote 2 down vote accepted

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

Recall that $$(1-x)^n = \sum_{k=0}^n \dbinom{n}k (-x)^k$$

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