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    <name>
      <text>9.2.7 (Coefficients of binomial expansions)</text>
    </name>
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      <text><![CDATA[<h3>Coefficients of binomial expansions</h3>

<p>We call expansions of the type we have seen above e.g. \((x+y)^8\) or \((x+y)^3\) etc <b>binomial expansions</b>. There is a reason for this. The end of the previous point, asked you to notice if you spotted anything about the coefficients of the expansions. The coefficients are given by the corresponding binomial coefficients. This is the conclusion of the Binomial Theorem which we will discuss in more detail in the following section. We will now focus on how to use this fact to more quickly calculate such expansions.</p>

<p>In general, if we wish to expand a bracket (called a binomial as it consists of two terms) of the form \((x+y)^n\) its expansion is given by 
\[
(x+y)^n = x^n + {n \choose 1} x^{n-1}y + {n \choose 2} x^{n-2}y^2 + \cdots + {n \choose r} x^{n-r}y^r + y^n.
\]</p>
<p>There are a few important notes we make about this calculation.<p>
<ol>
<li>If the power in the binomial expansion is \(n\), then there are \((n+1)\) terms in the expansion.</li>
<li>The sum of the powers of \(x\) and \(y\) in each term is equal to the power of the binomial i.e it is equal to \(n\).</li>
<li>The binomial coefficients can be found using Pascals triangle, in some cases this is the faster method, but for larger \(n\) we use the formula for the binomial coefficient we saw in the previous section.</li>
</ol>

<p>Let us see how to use this fact to calculate different expansions.</p>
<br/>
<h3>Example</h3>

<p>Calculate the following expansions.</p>
<ol type="a">
<li>(x+y)^6</li>
<li>(x+2y)^3</li>
<li>(3-2x)^5</li>
</ol>

<h3>Solution</h3>

<ol type="a">
<li>We can use the result above in each case. In this case \(n=6\) so the coefficients of each term are given by \({6\choose r}\) for \(r=1,2,3,4,5\), therefore the expansion is 
\[
(x+y)^6 = x^6 + {6 \choose 1} x^5y + {6 \choose 2} x^4y^2 + {6 \choose 3} x^3y^3 + {6 \choose 4} x^2y^4 + {6 \choose 5} xy^5 + y^6
\]
or in other words,
\[
(x+y)^6 =  x^6 + 6 x^5y + 15 x^4y^2 + 20 x^3y^3 + 15 x^2y^4 + 6 xy^5 + y^6
\]
</li>
<li>We can use the same method here, but we need to be carful, as this time there is a factor of 2 in front of the \(y\) term. In this case \(n = 3\), so the coefficients of each term are going to be \({6\choose r}\) for \(r=1,2\) however each coefficient will be multiplied by the factor of \(2\) from the \(y\) term. Thus,
\[
(x+2y)^3 = x^3 + {3 \choose 1} x^2 (2y) + {3 \choose 2} x (2y)^2 + (2y)^3,
\]
or in other words,
\[
(x+2y)^3 = x^3 + 6 x^2 y + 12x y^2 + 8y^3.
\]</li>
<li>We use the same method again, but keep track of the coefficients carefully. Also, notice that this time there is a minus sign in the binomial, so we would expect the signs of each term in the expansion to alternate. Here \(n = 5\) so there will be six terms, thus
\[
(3-2x)^5 = 3^5 + {5 \choose 1} 3^4 (-2x) + {5 \choose 2} 3^3 (-2x)^2 + {5 \choose 3} 3^2 (-2x)^3 + {5 \choose 4} 3^1 (-2x)^4 + (-2x)^5,
\] 
or in other words
\[
(3-2x)^5 = 243 - 810x +1080x^2 - 720x^3 +240x^4 -32x^5
\]</li>
</ol>]]></text>
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