MCR3U: Pascal's Triangle and the Binomial Theorem

Multiplying out $(x + 2)^4$ by hand takes four rounds of brackets. Pascal's triangle gives the coefficients of any power of a binomial at once, and the binomial theorem says where the powers go, so the whole expansion takes one line.

How It Works

Pascal's triangle. Start with $1$ at the top, which is row $0$ . Each new row starts and ends with $1$ , and every entry inside is the sum of the two entries above it. Row $4$ is $1, 4, 6, 4, 1$ and row $5$ is $1, 5, 10, 10, 5, 1$ . Every row reads the same backward, and the entries of row $n$ add to $2^n$ .

Expanding. The coefficients of $(a + b)^n$ are row $n$ of the triangle. The power of $a$ starts at $n$ and goes down by one each term, the power of $b$ starts at $0$ and goes up by one, and in every term the two powers add to $n$ . So $(a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$ .

Real binomials. In $(2x - 3)^n$ the first term is $2x$ and the second is $-3$ . Put each one in brackets before raising it to a power, so its coefficient and its sign come along: $(2x)^2 = 4x^2$ and $(-3)^3 = -27$ . A negative second term makes the signs alternate.

One term on its own. Write $\binom{n}{k}$ for entry $k$ of row $n$ , counting from $0$ . Then term $k + 1$ of $(a + b)^n$ is $\binom{n}{k}a^{n - k}b^k$ . To find the term with a particular power of $x$ , set the power in this general term equal to the one you want, solve for $k$ , and evaluate just that term.

Check. Substitute $x = 1$ into the binomial and into your expansion. The two must agree, which catches most dropped signs and coefficients.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Write row $6$ of Pascal's triangle.
Show solution
1. Start from row $5$ : $1, 5, 10, 10, 5, 1$ .
2. Add neighbours: $1 + 5 = 6$ , $5 + 10 = 15$ , $10 + 10 = 20$ , then the row mirrors itself, with $1$ at each end.
3. Check: $1 + 6 + 15 + 20 + 15 + 6 + 1 = 64 = 2^6$ .
Answer: $1, 6, 15, 20, 15, 6, 1$
Easy
Expand $(x + 2)^4$ .
Show solution
1. Row $4$ : $1, 4, 6, 4, 1$ .
2. $x^4 + 4x^3(2) + 6x^2(2)^2 + 4x(2)^3 + 2^4$ .
3. Check at $x = 1$ : $3^4 = 81$ and $1 + 8 + 24 + 32 + 16 = 81$ .
Answer: $x^4 + 8x^3 + 24x^2 + 32x + 16$
Medium
Expand $(2x - 3)^3$ .
Show solution
1. Row $3$ : $1, 3, 3, 1$ , with first term $2x$ and second term $-3$ .
2. $(2x)^3 + 3(2x)^2(-3) + 3(2x)(-3)^2 + (-3)^3$ .
3. $= 8x^3 + 3(4x^2)(-3) + 3(2x)(9) - 27$ .
4. Check at $x = 1$ : $(2 - 3)^3 = -1$ and $8 - 36 + 54 - 27 = -1$ .
Answer: $8x^3 - 36x^2 + 54x - 27$
Medium
Find the term in $x^3$ in the expansion of $(x + 3)^7$ .
Show solution
1. The general term is $\binom{7}{k}x^{7 - k}3^k$ . For $x^3$ , $7 - k = 3$ , so $k = 4$ .
2. Row $7$ is $1, 7, 21, 35, 35, 21, 7, 1$ , so $\binom{7}{4} = 35$ .
3. The term is $35x^3(3)^4 = 35(81)x^3$ .
Answer: $2835x^3$
Test level
Find the constant term in the expansion of $\left(x^2 - \frac{2}{x}\right)^6$ .
Show solution
1. The general term is $\binom{6}{k}(x^2)^{6 - k}\left(-\frac{2}{x}\right)^k = \binom{6}{k}(-2)^k x^{12 - 2k - k} = \binom{6}{k}(-2)^k x^{12 - 3k}$ .
2. A constant term has $x^0$ : $12 - 3k = 0$ , so $k = 4$ .
3. Row $6$ is $1, 6, 15, 20, 15, 6, 1$ , so $\binom{6}{4} = 15$ .
4. The term is $15(-2)^4 = 15(16)$ .
Answer: $240$

Learn It on Martian Lab

Martian Lab teaches this in the Pascal's Triangle & Binomial Expansion lesson of its MCR3U Discrete Functions unit. Watch it, then take the Discrete Functions practice tests to see if it stuck.

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MCR3U: Pascal's Triangle and the Binomial Theorem

How It Works

5 Worked Examples

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