In general the binomial theorem says: \left(a+b\right)^n=\binom{n}{0}a^nb^0+\binom{n}{1}a^{n-1}b^1+\ldots +\binom{n}{r}a^{n-r}b^{r}+\ldots +\binom{n}{n-1}a^1b^{n-1}+\binom{n}{n}a^0b^n

For example: \left(4x+1\right)^3=\binom{3}{0}\left(4x\right)^3\left(1\right)^0+\binom{3}{1}\left(4x\right)^{2}\left(1\right)^1+\binom{3}{2}\left(4x\right)^{1}\left(1\right)^{2} +\binom{3}{3}\left(4x\right)^0 \left(1\right)^{3}

The elements of Pascal's triangle can also be used to evaluate the coefficients as row n will given the coefficients of the terms of \left(a+b\right)^n in order.

Pascal's triangle with each row next to the corresponding (a+b)^n. The first row is just 1 corresponding to the coefficients of (a+b)^0, the second row is 1 1 corresponding to the coefficients of (a+b)^1, the third row is 1 2 1 corresponding to the coefficients of (a+b)^2, the fourth row is 1 3 3 1 corresponding to the coefficients of (a+b)^3, etc. The final row is 1 7 21 35 35 21 7 1 which corresponds to the coefficients of (a+b)^7.

Worked examples

Example 1

Evaluate each expression using the given approach.

Evaluate \binom{6}{3} using Pascal's triangle.

Approach

In Pascal's triangle, \binom{n}{r} is the entry in position r in row n. We start counting the rows and entries with 0.

Solution

We have n=6 and r=3. The entries in the first six rows of Pascal's triangle are as follows:

The 6th row is 1,\,6,\, 15,\, 20,\, 15,\, 6,\,1. Since the count for r begins at 0, \binom{6}{3} represents the 4th entry in the row. So, \binom{6}{3}=20.

Reflection

We can check our answer by using the factorial expansion of \binom{6}{3}.

\displaystyle \binom{n}{r}	\displaystyle =	\displaystyle \dfrac{n!}{(n-r)!r!}	Formula
\displaystyle \binom{6}{3}	\displaystyle =	\displaystyle \dfrac{6!}{(6-3)!3!}	Substitute n=6 and r=3
	\displaystyle =	\displaystyle \dfrac{6!}{3!3!}	Evaluate the difference
	\displaystyle =	\displaystyle 28	Evaluate the factorials

Evaluate {}_{7}C_{4} using formula.

Approach

The formula is {}_{n}C_{r}=\dfrac{n!}{(n-r)!r!}.

Solution

We have n=7 and r=4.

\displaystyle {}_{n}C_{r}	\displaystyle =	\displaystyle \dfrac{n!}{(n-r)!r!}	Formula
\displaystyle {}_{7}C_{4}	\displaystyle =	\displaystyle \dfrac{7!}{(7-4)!4!}	Substitute n=7 and r=4
	\displaystyle =	\displaystyle \dfrac{7!}{3!4!}	Evaluate the difference
	\displaystyle =	\displaystyle \dfrac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{(3\cdot2\cdot1)(4\cdot3\cdot2\cdot1)}	Write out the expansion of the factorials
	\displaystyle =	\displaystyle \dfrac{7\cdot5}{1}	Reduce the fraction
	\displaystyle =	\displaystyle 35	Evaluate the product

Reflection

Finding row 7 of Pascal's triangle would be time consuming. We can also check this using a calculator by using the button marked with {}_{n}C_{r} or \text{nCr} which may require the use of the "shift" button. The calculator should confirm that {}_{7}C_{4}=35.

Example 2

Find the middle term in the expansion \left( 2x^{3} - 3y^{2}\right)^{8}.

Approach

Each term in the expansion (a+b)^n is of the form {}_{n}C_{r}\cdot a^{n-r} \cdot b^r.

Since n=8 and we are looking for the middle term, we will use n=8, r=4.

Solution

The expression \left( 2x^{3} - 3y^{2}\right)^{8} can be rewritten as \left( 2x^{3} +\left(- 3y^{2}\right)\right)^{8}.

For (a+b)^n={}_{n}C_{r}\cdot a^{n-r} \cdot b^r, we have:

a=2x^{3}
b=- 3y^{2}
n=8
r=4

\displaystyle {}_{n}C_{r}\cdot a^{n-r} \cdot b^r	\displaystyle =	\displaystyle {}_{8}C_{4}\cdot \left(2x^3\right)^{8-4} \cdot \left(-3y^2\right)^4	Substitution
	\displaystyle =	\displaystyle \dfrac{8!}{(8-4)!4!} \cdot \left(2x^3\right)^{8-4} \cdot \left(-3y^2\right)^4	Rewrite {}_{8}C_{4} in factorial form
	\displaystyle =	\displaystyle \dfrac{8!}{4!4!} \cdot \left(2x^3\right)^{4} \cdot \left(-3y^2\right)^4	Evaluate the difference
	\displaystyle =	\displaystyle 70 \cdot \left(2x^3\right)^{4} \cdot \left(-3y^2\right)^4	Evaluate the factorials
	\displaystyle =	\displaystyle 70 \cdot 16x^{12} \cdot 81y^8	Evaluate the powers
	\displaystyle =	\displaystyle 90\,720x^{12}y^{8}	Simplify

Reflection

An alternative solution would be to use Pascal's Triangle to get the row for the power of 8: 1 \quad 8 \quad 28 \quad 56 \quad 70 \quad 56 \quad 28 \quad 8 \quad 1

Then identifying that the middle term will use the 70, as it is in the middle.

Counting down from 8 for the exponent on the first term of the binomial and counting up from 0 for the exponent on the second term of the binomial, we get that they both have exponent 4, giving us:70 \cdot \left(2x^3\right)^4 \left(-3y^2\right)^4=90\,720x^{12}y^{8}

Example 3

Use the binomial theorem to expand \left(p+4\right)^{5}.

Approach

Applying the binomial theorem, each term in the expansion (a+b)^n is of the form {}_{n}C_{r}\cdot a^{n-r} \cdot b^r where n is the power of the binomial and r is the value corresponding to the (r+1)th term. The expansion is given as follows:

(a+b)^n={}_{n}C_{0}\cdot a^{n-0} \cdot b^0 + {}_{n}C_{1}\cdot a^{n-1} \cdot b^1 + {}_{n}C_{2}\cdot a^{n-2} \cdot b^2 + \ldots + {}_{n}C_{n}\cdot a^{n-n} \cdot b^n

Solution

We have n=5 and r is an integer value from 0 to 5.

The coefficients will be the values found in row 5 in Pascal's triangle which are: 1, \, 5,\, 10,\, 10,\, 5,\, 1

\displaystyle \left(p+4\right)^{5}	\displaystyle =	\displaystyle {}_{5}C_{0}\cdot p^{5-0} \cdot 4^0 +{}_{5}C_{1}\cdot p^{5-1} \cdot 4^1+{}_{5}C_{2}\cdot p^{5-2} \cdot 4^2 \\ +{}_{5}C_{3}\cdot p^{5-3} \cdot 4^3 +{}_{5}C_{4}\cdot p^{5-4} \cdot 4^4+{}_{5}C_{5}\cdot p^{5-5} \cdot 4^5
	\displaystyle =	\displaystyle 1\cdot p^{5-0} \cdot 4^0 +5\cdot p^{5-1} \cdot 4^1+10\cdot p^{5-2} \cdot 4^2 \\ +10\cdot p^{5-3} \cdot 4^3 +5\cdot p^{5-4} \cdot 4^4+1\cdot p^{5-5} \cdot 4^5
	\displaystyle =	\displaystyle 1\cdot p^{5} \cdot 4^0 +5\cdot p^{4} \cdot 4^1+10\cdot p^{3} \cdot 4^2 \\+10\cdot p^{2} \cdot 4^3 +5\cdot p^{1} \cdot 4^4+1\cdot p^{0} \cdot 4^5
	\displaystyle =	\displaystyle p^{5} +20p^{4}+160p^{3}+640p^{2}+1280p+1024

Reflection

If we use the formula instead of Pascal's triangle, we get the same answer:

\displaystyle \left(p+4\right)^{5}	\displaystyle =	\displaystyle {}_{5}C_{0}\cdot p^{5-0} \cdot 4^0 +{}_{5}C_{1}\cdot p^{5-1} \cdot 4^1+{}_{5}C_{2}\cdot p^{5-2} \cdot 4^2 \\ +{}_{5}C_{3}\cdot p^{5-3} \cdot 4^3 +{}_{5}C_{4}\cdot p^{5-4} \cdot 4^4+{}_{5}C_{5}\cdot p^{5-5} \cdot 4^5	Expansion
	\displaystyle =	\displaystyle \binom{5}{0}\cdot p^{5-0} \cdot 4^0 +\binom{5}{1}\cdot p^{5-1} \cdot 4^1+\binom{5}{2}\cdot p^{5-2} \cdot 4^2 \\ +\binom{5}{3}\cdot p^{5-3} \cdot 4^3 +\binom{5}{4}\cdot p^{5-4} \cdot 4^4+\binom{5}{5}\cdot p^{5-5} \cdot 4^5	Substitute {}_{n}C_{r}=\binom{n}{r}
	\displaystyle =	\displaystyle \dfrac{5!}{(5-0)!0!}\cdot p^{5-0} \cdot 4^0 +\dfrac{5!}{(5-1)!1!}\cdot p^{5-1} \cdot 4^1\\+\dfrac{5!}{(5-2)!2!}\cdot p^{5-2} \cdot 4^2 +\dfrac{5!}{(5-3)!3!}\cdot p^{5-3} \cdot 4^3 \\+\dfrac{5!}{(5-4)!4!}\cdot p^{5-4} \cdot 4^4+\dfrac{5!}{(5-5)!5!}\cdot p^{5-5} \cdot 4^5	Rewrite \binom{n}{r} in factorial form
	\displaystyle =	\displaystyle \dfrac{5!}{5!0!}\cdot p^{5} \cdot 4^0 +\dfrac{5!}{4!1!}\cdot p^{4} \cdot 4^1+\dfrac{5!}{3!2!}\cdot p^{3} \cdot 4^2 \\+\dfrac{5!}{2!3!}\cdot p^{2} \cdot 4^3 +\dfrac{5!}{1!4!}\cdot p^{1} \cdot 4^4+\dfrac{5!}{0!5!}\cdot p^{0} \cdot 4^5	Evaluate the difference
	\displaystyle =	\displaystyle 1\cdot p^{5} \cdot 4^0 +5\cdot p^{4} \cdot 4^1+10\cdot p^{3} \cdot 4^2 \\+10\cdot p^{2} \cdot 4^3 +5\cdot p^{1} \cdot 4^4+1\cdot p^{0} \cdot 4^5	Evaluate the factorials
	\displaystyle =	\displaystyle p^{5} +20p^{4}+160p^{3}+640p^{2}+1280p+1024	Simplify

Honors: 3.07 Binomial expansion

Concept summary

Worked examples

Example 1

Approach

Solution

Reflection

Approach

Solution

Reflection

Example 2

Approach

Solution

Reflection

Example 3

Approach

Solution

Reflection

Outcomes

MA.912.AR.1.11

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