The Binomial Theorem Solutions

• When a binomial expression is raised to the $n^{th}$ power where n is a whole number,    patterns emerge    in the coefficients and the powers of the terms.

Note

The first term is raised to the $n^{th}$ power and decreases until reaching zero. The second term is raised to the zero power and increases until reaching the $n^{th}$ power. The coefficients are the same as those in the $n^{th}$ row of Pascal’s triangle.

• The Binomial Theorem determines the coefficients of an $n^{th}$ power binomial    in expanded form when n is a natural number   .

• In this formula, the “stacked” terms inside the parentheses represent    a combination, nCr   .

Note

Stacking the numbers is another notation for writing combinations.

• You can use Pascal’s Triangle in place of the Binomial Theorem, but you would need to    create it with n rows   .

Note

The math shorthand for the Binomial Theorem uses the summation symbol, $\sum$. However, we will use the expanded binomial with combinations because the summation symbol is not commonly used in this unit.

• The two terms of the binomial expression are    x and y   .
• If x or y has a    coefficient   , use    parentheses    to correctly expand the binomial.

• When the binomial contains a numerical term:

  •    Raise the term to the correct power   
  •    Multiply by the correct combination   
• This means the    numbers will not always equal the combination    once the terms are completely simplified.
• When the terms of a binomial expression are subtracted, ${(x–y)}^n$, the signs of the terms will    alternate between positive and negative   , starting with a positive value.
• Finding the $n^{th}$ term:

  • The Binomial Theorem can be used to find a specific term in a binomial expression    without expanding it fully   .

Example 3

Expand the binomial with Pascal’s Triangle.

${\left(x+y\right)}^5$

Row 5:  $1 5 10 10 5 1$  

$1x^5{y}^0+5x^4{y}^1+10x^3{y}^2+10x^2{y}^3+5x^1{y}^4+1x^0{y}^5$

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

Example 4

Determine the middle term of the binomial.

Example 5

Determine the third term of the binomial expression.

${\left(3x+y\right)}^7$

Note

The sum of the exponents must equal the degree.

Example 6

Expand the binomial.

${\left(2x–\frac{1}{3}y\right)}^4$

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

$= \left(1\right)\left(2^4\right){\left(–\frac{1}{3}\right)}^0{x}^4{y}^0+ \left(4\right)\left(2^3\right){\left(–\frac{1}{3}\right)}^1{x}^3{y}^1+ \left(6\right)\left(2^2\right){\left(–\frac{1}{3}\right)}^2{x}^2{y}^2+$$\left(4\right){\left(2\right)}^1{(–\frac{1}{3})}^3{x}^1{y}^3+\left(1\right){\left(2\right)}^0{(–\frac{1}{3})}^4{x}^0{y}^4$

$=16x^4–\frac{32}{3}{x}^{3}y+\frac{24}{9}{x}^2{y}^2–\frac{8}{27}x{y}^3+\frac{1}{81}{y}^4$