Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

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 $(-2)^n$.

share|improve this answer
    
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 $$(1-x)^n = \sum_{k=0}^n \dbinom{n}k (-x)^k$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.