# Manipulating Sums

• Oct 21st 2010, 11:59 AM
evanator
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 $\displaystyle\sum_{r=0}^n ^nC_r 2^r = 3^n$

So, we already know that $\displaystyle\sum_{r=0}^n ^nC_r = 2^n$, because

$(a+b)^n = \displaystyle\sum_{r=0}^n ^nC_r a^{n-r} b^{r}$

Letting $a = 1$ and $b = 1$:

$(1+1)^n = \displaystyle\sum_{r=0}^n ^nC_r 1^{n-r} 1^{r}$

$2n = \displaystyle\sum_{r=0}^n ^nC_r$

Getting back to the matter at hand, we can take out the $2^n$:

$2^n \displaystyle\sum_{r=0}^n 2^r$

We can also write it as:

$\displaystyle\sum_{r=0}^n 2^{n+r}$

This gives the same series, because

$(2^n \times 2^0) + (2^n \times 2^1) + (2^n \times 2^2) + ... + (2^n \times 2^n)$

is the same as

$2^{n+0} + 2^{n +1} + 2^{n+2} + ... + 2^{2n}$

I don't know what to do to get the $3^n$ though.
Any help would be appreciated.

Regards,

Evanator
• Oct 21st 2010, 12:01 PM
Plato
Notice that $3^n=(1+2)^n$. Just expand.
• Oct 21st 2010, 12:15 PM
evanator
Looks like I couldn't see the wood for the trees. Thank you, Plato.
Expanding out, bearing in mind that $(1+2)^n = 3^n$:

$(a+b)^n = \displaystyle\sum_{r=0}^n ^nC_r a^{n-r} b^r$

$(1+2)^n = \displaystyle\sum_{r=0}^n ^nC_r 1^{n-r} 2^r$

$3^n = \displaystyle\sum_{r=0}^n ^nC_r 2^r$