
Manipulating Sums
Hi there,
I am trying to solve the following problem. I have gotten a fair way into it, but the last steps elude me. Here is the problem:
Show that $\displaystyle \displaystyle\sum_{r=0}^n ^nC_r 2^r = 3^n$
So, we already know that $\displaystyle \displaystyle\sum_{r=0}^n ^nC_r = 2^n$, because
$\displaystyle (a+b)^n = \displaystyle\sum_{r=0}^n ^nC_r a^{nr} b^{r}$
Letting $\displaystyle a = 1$ and $\displaystyle b = 1$:
$\displaystyle (1+1)^n = \displaystyle\sum_{r=0}^n ^nC_r 1^{nr} 1^{r}$
$\displaystyle 2n = \displaystyle\sum_{r=0}^n ^nC_r$
Getting back to the matter at hand, we can take out the $\displaystyle 2^n$:
$\displaystyle 2^n \displaystyle\sum_{r=0}^n 2^r$
We can also write it as:
$\displaystyle \displaystyle\sum_{r=0}^n 2^{n+r}$
This gives the same series, because
$\displaystyle (2^n \times 2^0) + (2^n \times 2^1) + (2^n \times 2^2) + ... + (2^n \times 2^n)$
is the same as
$\displaystyle 2^{n+0} + 2^{n +1} + 2^{n+2} + ... + 2^{2n}$
I don't know what to do to get the $\displaystyle 3^n$ though.
Any help would be appreciated.
Regards,
Evanator

Notice that $\displaystyle 3^n=(1+2)^n$. Just expand.

Looks like I couldn't see the wood for the trees. Thank you, Plato.
Expanding out, bearing in mind that $\displaystyle (1+2)^n = 3^n$:
$\displaystyle (a+b)^n = \displaystyle\sum_{r=0}^n ^nC_r a^{nr} b^r$
$\displaystyle (1+2)^n = \displaystyle\sum_{r=0}^n ^nC_r 1^{nr} 2^r$
$\displaystyle 3^n = \displaystyle\sum_{r=0}^n ^nC_r 2^r$