# The Easy Way To Find $(x+y)^n$

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In many courses in Mathematics we oftentimes have to expand out things like $(x+y)^n$ where $n$ is an integer and $x$ and $y$ are real numbers. So many of us will happily expand it out and collect like terms. But this is kind of hard. Isn't it? Well we want to show you an easier and shockingly fast way to just do it.

First, let's do the actual Algebra to show how much of a pain in the butt this is:
\begin{eqnarray}
(x+y)^3&=&(x+y)(x+y)(x+y)\\
&=&(x^2+xy+xy+y^2)(x+y)\\
&=&(x^2+2xy+y^2)(x+y)\\
&=&x^3+x^2y+2x^2y+2xy^2+xy^2+y^3\\
&=&x^3+3x^2y+3xy^2+y^3.
\end{eqnarray}

The quick way to do this is with the formula
$(x+y)^n=\sum_{j=0}^nC_nx^{n-j}y^j.$
The symbol $\sum$ is the summation formula that tells is to add from $0$ to $n$ and as we go we will have a number $C_n$ multiplied by $x^{n-j}y^j$. In other words,
$(x+y)^n=C_0x^ny^0+C_1x^{n-1}y^1+\cdots+C_{n-1}xy^{n-1}+C_nx^0y^n.$
The numbers $C_0,\,C_1,\,\cdots$ are referred to as Binomial Coefficients and finding them is the hard part in this formula.

It turns out for $n\leq 4$ finding the Binomial Coefficients ($C_n$'s) is really easy with a simple trick. The trick is to find the number $11^n$. (We recommend using a calculator since you want to get this done quickly.) It turns out that the digits in $11^n$ are the coefficients $C_n$! In our example above $n=3$. So on your calculator you find that $11^3=1331$. So this means that $C_0=1$, $C_1=3$, $C_2=3$, $C_3=1$. So now using the formula we have
$(x+y)^3=C_0x^3y^0+C_1x^2y+C_2xy^2+C_3x^0y^3.$Now since $x^0=y^0=1$ and because we found our coefficients from $11^3$ we have
$(x+y)^3=x^3+3x^2y+3xy^2+y^3.$

So now if you have to find $(x+5)^4$ you will first find that $11^4=14641$ and that
\begin{eqnarray}
(x+5)^4&=&x^4+4x^35^1+6x^25^2+4x5^3+5^4\\
&=&x^4+20x^3+150x^2+500x+625.
\end{eqnarray}

We want to emphasize that this trick is only good for $n$ less than or equal to four. But because $(x+y)^n$ shows up so much when $n\leq 4$ it turns that this is a very useful thing to keep in mind. Finally, this trick is rooted in some pretty deep Mathematics related to Pascal's Triangle, Factorials, and the Binomial Theorem. But that's a story for another day.