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Binomial Theorem | ${(x+y)}^{n}={\displaystyle \sum _{k-0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right)}{x}^{n-k}{y}^{k}$ |
$(r+1)th\text{\hspace{0.17em}}$ term of a binomial expansion | $$\left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}$$ |
What is a binomial coefficient, and how it is calculated?
A binomial coefficient is an alternative way of denoting the combination $\text{\hspace{0.17em}}C(n,r).\text{\hspace{0.17em}}$ It is defined as $\text{\hspace{0.17em}}\left(\begin{array}{c}n\\ r\end{array}\right)=\text{\hspace{0.17em}}C(n,r)\text{\hspace{0.17em}}=\frac{n!}{r!(n-r)!}.$
What role do binomial coefficients play in a binomial expansion? Are they restricted to any type of number?
What is the Binomial Theorem and what is its use?
The Binomial Theorem is defined as $\text{\hspace{0.17em}}{(x+y)}^{n}={\displaystyle \sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right){x}^{n-k}}{y}^{k}\text{\hspace{0.17em}}$ and can be used to expand any binomial.
When is it an advantage to use the Binomial Theorem? Explain.
For the following exercises, evaluate the binomial coefficient.
$\left(\begin{array}{c}5\\ 3\end{array}\right)$
$\left(\begin{array}{c}9\\ 7\end{array}\right)$
$\left(\begin{array}{c}25\\ 11\end{array}\right)$
$\left(\begin{array}{c}200\\ 199\end{array}\right)$
For the following exercises, use the Binomial Theorem to expand each binomial.
${(5a+2)}^{3}$
${(2x+3y)}^{4}$
${(4x+2y)}^{5}$
$1024{x}^{5}+2560{x}^{4}y+2560{x}^{3}{y}^{2}+1280{x}^{2}{y}^{3}+320x{y}^{4}+32{y}^{5}$
${(3x-2y)}^{4}$
${(4x-3y)}^{5}$
$1024{x}^{5}-3840{x}^{4}y+5760{x}^{3}{y}^{2}-4320{x}^{2}{y}^{3}+1620x{y}^{4}-243{y}^{5}$
${\left(\frac{1}{x}+3y\right)}^{5}$
${({x}^{-1}+2{y}^{-1})}^{4}$
$\frac{1}{{x}^{4}}+\frac{8}{{x}^{3}y}+\frac{24}{{x}^{2}{y}^{2}}+\frac{32}{x{y}^{3}}+\frac{16}{{y}^{4}}$
${(\sqrt{x}-\sqrt{y})}^{5}$
For the following exercises, use the Binomial Theorem to write the first three terms of each binomial.
${(x-1)}^{18}$
${(x-2y)}^{8}$
${(3a+b)}^{20}$
$3,486,784,401{a}^{20}+23,245,229,340{a}^{19}b+73,609,892,910{a}^{18}{b}^{2}$
${(2a+4b)}^{7}$
${({x}^{3}-\sqrt{y})}^{8}$
${x}^{24}-8{x}^{21}\sqrt{y}+28{x}^{18}y$
For the following exercises, find the indicated term of each binomial without fully expanding the binomial.
The fourth term of $\text{\hspace{0.17em}}{(2x-3y)}^{4}$
The fourth term of $\text{\hspace{0.17em}}{(3x-2y)}^{5}$
$-720{x}^{2}{y}^{3}$
The third term of $\text{\hspace{0.17em}}{(6x-3y)}^{7}$
The eighth term of $\text{\hspace{0.17em}}{(7+5y)}^{14}$
$220,812,466,875,000{y}^{7}$
The seventh term of $\text{\hspace{0.17em}}{(a+b)}^{11}$
The fifth term of $\text{\hspace{0.17em}}{(x-y)}^{7}$
$35{x}^{3}{y}^{4}$
The tenth term of $\text{\hspace{0.17em}}{(x-1)}^{12}$
The ninth term of $\text{\hspace{0.17em}}{(a-3{b}^{2})}^{11}$
$1,082,565{a}^{3}{b}^{16}$
The fourth term of $\text{\hspace{0.17em}}{\left({x}^{3}-\frac{1}{2}\right)}^{10}$
The eighth term of $\text{\hspace{0.17em}}{\left(\frac{y}{2}+\frac{2}{x}\right)}^{9}$
$\frac{1152{y}^{2}}{{x}^{7}}$
For the following exercises, use the Binomial Theorem to expand the binomial $f(x)={(x+3)}^{4}.$ Then find and graph each indicated sum on one set of axes.
Find and graph $\text{\hspace{0.17em}}{f}_{1}(x),\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}{f}_{1}(x)\text{\hspace{0.17em}}$ is the first term of the expansion.
Find and graph $\text{\hspace{0.17em}}{f}_{2}(x),\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}{f}_{2}(x)\text{\hspace{0.17em}}$ is the sum of the first two terms of the expansion.
${f}_{2}(x)={x}^{4}+12{x}^{3}$
Find and graph $\text{\hspace{0.17em}}{f}_{3}(x),\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}{f}_{3}(x)\text{\hspace{0.17em}}$ is the sum of the first three terms of the expansion.
Find and graph $\text{\hspace{0.17em}}{f}_{4}(x),\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}{f}_{4}(x)\text{\hspace{0.17em}}$ is the sum of the first four terms of the expansion.
${f}_{4}(x)={x}^{4}+12{x}^{3}+54{x}^{2}+108x$
Find and graph $\text{\hspace{0.17em}}{f}_{5}(x),\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}{f}_{5}(x)\text{\hspace{0.17em}}$ is the sum of the first five terms of the expansion.
In the expansion of $\text{\hspace{0.17em}}{(5x+3y)}^{n},\text{\hspace{0.17em}}$ each term has the form $\text{\hspace{0.17em}}\left(\begin{array}{c}n\\ k\end{array}\right){a}^{n\u2013k}{b}^{k},\text{where}k\text{\hspace{0.17em}}$ successively takes on the value $\text{\hspace{0.17em}}0,1,2,\text{\hspace{0.17em}}\mathrm{...},\text{\hspace{0.17em}}n.$ If $\text{\hspace{0.17em}}\left(\begin{array}{c}n\\ k\end{array}\right)=\left(\begin{array}{c}7\\ 2\end{array}\right),\text{\hspace{0.17em}}$ what is the corresponding term?
$590,625{x}^{5}{y}^{2}$
In the expansion of $\text{\hspace{0.17em}}{\left(a+b\right)}^{n},\text{\hspace{0.17em}}$ the coefficient of $\text{\hspace{0.17em}}{a}^{n-k}{b}^{k}\text{\hspace{0.17em}}$ is the same as the coefficient of which other term?
Consider the expansion of $\text{\hspace{0.17em}}{(x+b)}^{40}.\text{\hspace{0.17em}}$ What is the exponent of $b$ in the $k\text{th}$ term?
$k-1$
Find $\text{\hspace{0.17em}}\left(\begin{array}{c}n\\ k-1\end{array}\right)+\left(\begin{array}{c}n\\ k\end{array}\right)\text{\hspace{0.17em}}$ and write the answer as a binomial coefficient in the form $\text{\hspace{0.17em}}\left(\begin{array}{c}n\\ k\end{array}\right).\text{\hspace{0.17em}}$ Prove it. Hint: Use the fact that, for any integer $\text{\hspace{0.17em}}p,\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}p\ge 1,\text{\hspace{0.17em}}p!=p(p-1)!\text{.}$
$\left(\begin{array}{c}n\\ k-1\end{array}\right)+\left(\begin{array}{l}n\\ k\end{array}\right)=\left(\begin{array}{c}n+1\\ k\end{array}\right);\text{\hspace{0.17em}}$ Proof:
$\begin{array}{}\\ \\ \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\begin{array}{c}n\\ k-1\end{array}\right)+\left(\begin{array}{l}n\\ k\end{array}\right)\\ =\frac{n!}{k!\left(n-k\right)!}+\frac{n!}{\left(k-1\right)!\left(n-\left(k-1\right)\right)!}\\ =\frac{n!}{k!(n-k)!}+\frac{n!}{\left(k-1\right)!\left(n-k+1\right)!}\\ =\frac{\left(n-k+1\right)n!}{\left(n-k+1\right)k!(n-k)!}+\frac{kn!}{k\left(k-1\right)!\left(n-k+1\right)!}\\ =\frac{\left(n-k+1\right)n!+kn!}{k!\left(n-k+1\right)!}\\ =\frac{\left(n+1\right)n!}{k!\left(\left(n+1\right)-k\right)!}\\ =\frac{\left(n+1\right)!}{k!\left(\left(n+1\right)-k\right)!}\\ =\left(\begin{array}{c}n+1\\ k\end{array}\right)\end{array}$
Which expression cannot be expanded using the Binomial Theorem? Explain.
The expression $\text{\hspace{0.17em}}{({x}^{3}+2{y}^{2}-z)}^{5}\text{\hspace{0.17em}}$ cannot be expanded using the Binomial Theorem because it cannot be rewritten as a binomial.
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