Question

Find the expansion of$\mathbf{(}\mathit{x}\mathbf{+}\mathit{y}{\mathbf{)}}^{\mathbf{5}}$

a) using combinatorial reasoning, as in Example 1.

b) using the binomial theorem.

Short answer

$x^5+5x^{4}y+10x^3{y}^2+10x^2{y}^3+5x{y}^4+y^5$

Step-by-step solution

Formula for Binomial theorem

Binomial theorem is:$\mathbf{(}\mathit{x}\mathbf{+}\mathit{y}{\mathbf{)}}^{\mathbf{n}}\mathbf{=}\mathbf{\sum }_{\mathbf{j}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\mathbf{ }\left(\begin{array}{l}n\\ j\end{array}\right){\mathit{x}}^{\mathbf{n}\mathbf{-}\mathbf{j}}{\mathit{y}}^{\mathbf{j}}$

Use Distributive property and simplify

(a)

Write the given statement as a product and then use the distributive property

$\begin{array}{r}(x+y{)}^4=(x+y\left)\right(x+y\left)\right(x+y\left)\right(x+y\left)\right(x+y)\\ (x+y{)}^4=\left(x^2+xy+xy+y^2\right)\left(x^2+xy+xy+y^2\right)(x+y)\end{array}$

$\begin{array}{r}(x+y{)}^4=\left(x^2+2xy+y^2\right)\left(x^2+2xy+y^2\right)(x+y)\\ (x+y{)}^4=\left(x^4+2x^{3}y+x^2{y}^2+2x^{3}y+4x^2{y}^2+2x{y}^3+x^2{y}^2+2x{y}^3+y^4\right)(x+y)\\ (x+y{)}^4=\left(x^4+4x^{3}y+6x^2{y}^2+4x{y}^3+y^4\right)(x+y)\end{array}$

role="math" localid="1668681910506" $\begin{array}{r}(x+y{)}^4=x^5+4x^{4}y+6x^3{y}^2+4x^2{y}^3+x{y}^4+x^{4}y+4x^3{y}^2+6x^2{y}^3+4x{y}^4+y^5\\ (x+y{)}^4=x^5+5x^{4}y+10x^3{y}^2+10x^2{y}^3+5x{y}^4+y^5\end{array}$

Use Binomial theorem and simplify

(b)

Use the binomial theorem:

$(x+y{)}^5=\left(\begin{array}{l}5\\ 0\end{array}\right)x^5{y}^0+\left(\begin{array}{l}5\\ 1\end{array}\right)x^4{y}^1+\left(\begin{array}{l}5\\ 2\end{array}\right)x^3{y}^2+\left(\begin{array}{l}5\\ 3\end{array}\right)x^2{y}^3+\left(\begin{array}{l}5\\ 4\end{array}\right)x^1{y}^4+\left(\begin{array}{l}5\\ 5\end{array}\right)x^0{y}^5$

$(x+y{)}^5=\frac{5!}{0!5!}{x}^5+\frac{5!}{1!4!}{x}^{4}y+\frac{5!}{2!3!}{x}^3{y}^2+\frac{5!}{3!2!}{x}^2{y}^3+\frac{5!}{4!1!}x{y}^4+\frac{5!}{5!0!}{y}^5$

$\begin{array}{r}(x+y{)}^5=1x^5+5x^{4}y+10x^3{y}^2+10x^2{y}^3+5x{y}^4+y^5\\ (x+y{)}^5=x^5+5x^{4}y+10x^3{y}^2+10x^2{y}^3+5x{y}^4+y^5\end{array}$