## Algebraic Expansion

If you expand the polynomial $\left (a+b\right )^2$, the resulting polynomial $a^2+2ab+b^2$ has three terms whose coefficients are 1, 2, and 1.  Find the sum of the coefficients of the terms in the expansion of $\left (a+b\right )^{10}$ .
Source: mathcontest.olemiss.edu 5/5/2008

SOLUTION
We use the “area” model to represent the product of polynomials. For example, $\left (a+b\right )\left (a+b\right )$ is thought of as $area=length\times width$ where $length=\left (a+b\right )$ and $width=\left (a+b\right )$.

The product $\left (a+b\right )^2=\left (a+b\right )\left (a+b\right )$ is calculated as follows:

 a b a a² ab b ab b²

Thus, $\left (a+b\right )^2=a^2+2ab+b^2$.
Sum of the coefficients: $1+2+1=4$.

The problem asks for the sum of the coefficients, not what the individual coefficients look like in the finished product so it is not necessary to combine the like terms before calculating the sum of the coefficients. Instead, we will leave all the like terms un-combined as follows:
$\left (a+b\right )^2=a^2+ab+ab+b^2$
. The sum of the coefficients is $1+1+1+1=4$

Similarly, the product $\left (a+b\right )^3=\left (a+b\right )\left (a^2+2ab+b^2\right )$ is calculated as follows:

 a² 2ab b² a a³ 2a²b ab² b a²b 2ab² b³

Thus, $\left (a+b\right )^3=a^3+3a^2b+3ab^2+b^3$.
Sum of the coefficients: $1+3+3+1=8$.

If we left the like terms un-combined, the same operation looks like this:

 a² ab ab b² a a³ a²b a²b ab² b a²b ab² ab² b³

All 8 coefficients are equal to 1. Their sum is $1+1+1+1+1+1+1+1=8$.

Better yet, let’s leave out all the variables and just use the coefficients. The same operation looks like this:

 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Let’s use this procedure and calculate the sum of the coefficients of $\left (a+b\right )^4$.

 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

The sum of the coefficients equals 16.

For verification, let’s calculate the product $\left (a+b\right )^4=\left (a+b\right )\left (a^3+3a^2b+3ab^2+b^3\right )$ the old way and see if we get the same result:

 a³ 3a²b 3ab² b³ a a⁴ 3a³b 3a²b² ab³ b a³b 3a²b² 3ab³ b⁴

Thus, $\left (a+b\right )^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4$.
Sum of the coefficients: $1+4+6+4+1=16$.

General formula
Let $S_i$ be the sum of the coefficients of $\left (a+b\right )^i$. Then,
$\left (a+b\right )^2: S_2=1+1+1+1=4=2^2$.
$\left (a+b\right )^3: S_3=2\times S_2=2\times 2^2=2^3$.
$\left (a+b\right )^4:=2\times S_3=2\times 2^3=2^4$.

In general, for $\left (a+b\right )^n$:
$S_n=2\times S_{n-1}$
$=2\times 2^{n-1}$
$=2^n$
.

Sum of coefficients of $\mathbf{\left (a+b\right )^{10}}$.
$S_{10}=2^{10}=1024$.

If you are familiar with the Pascal Triangle, you can derive the individual coefficients of $\left (a+b\right )^{10}$ and calculate their sum as follows:
$1+10+45+120+210+252+210+120+45+10+1=1024$.