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Find

$$\sum_{r=0}^n3^r\binom{n}r$$

When I tried to do it I got

$$1 + 3n + 3^2\cdot \frac{n(n-1)}2 + \ldots + 3^{n-1}\cdot n + 3^n$$

but I couldn't equate it to $4^n$ which is the answer I got by substituting $n$ for various numbers.

Can you tell me the correct mathematical way of getting the answer?

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n sigma r=0 3^r nCr –  chndn Dec 9 '12 at 9:00
    
I took the liberty of changing your $_nC_r$ notation to the preferable $\binom{n}r$. –  Brian M. Scott Dec 9 '12 at 9:01
    
Thanks for the edit –  chndn Dec 9 '12 at 9:03
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1 Answer

up vote 4 down vote accepted

Just use the binomial theorem. It says that

$$(x+y)^n=\sum_{r=0}^n\binom{n}rx^ry^{n-r}\;;\tag{1}$$

now let $x=3$ and $y=1$, and see that the righthand side of $(1)$ is exactly your sum, while the lefthand side is indeed $4^n$.

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