Study notes

Binomial Expansion

Expanding $(a+b)^n$ for a positive integer $n$ using $\binom{n}{r}$, and finding a specific term or coefficient without expanding fully.

Learn it step by step

$\binom{n}{r}$ counts a choice, and it's the coefficient engine of the expansion

$(r n) = \frac{n !}{r ! ( n - r )!}$ is read "n choose r". For example $(2 7) = \frac{7 !}{2 ! \times 5 !} = \frac{7 \times 6}{2 \times 1} = 21$ . These numbers are exactly the coefficients that appear when you expand $(a + b)^{n}$ .

The general term formula finds any term without expanding everything

For $(a + b)^{n}$ , the term containing $b^{r}$ is $T_{r + 1} = (r n) a^{n - r} b^{r}$ , where $r$ runs from 0 to $n$ . The full expansion has exactly $n + 1$ terms.

The coefficients are symmetric: $\binom{n}{0} = \binom{n}{n} = 1$

$(0 n)$ and $(n n)$ are always 1, since there is only one way to pick none or all of the $n$ factors. More generally $(r n) = (n - r n)$ , so the row of coefficients reads the same forwards and backwards.

A negative second term makes the signs alternate

Rewrite $(a - b)^{n}$ as $(a + (- b))^{n}$ , so the general term becomes $(r n) a^{n - r} (- b)^{r}$ , which carries a factor of $(- 1)^{r}$ . The sign flips from term to term: positive when $r$ is even, negative when $r$ is odd.

To find a specific coefficient, match the power of $x$ to the right $r$, then substitute

Write the general term for the expression, work out which power of $x$ it produces in terms of $r$ , set that equal to the power you want, and solve for $r$ . Then plug that single value of $r$ back in; there is no need to expand every other term.

A term "independent of $x$" is the one whose total power of $x$ is zero

When an expression mixes $x$ and $1/ x$ (or similar), the power of $x$ in the general term is some expression in $r$ . Set that expression equal to 0 and solve for $r$ to find which term has no $x$ in it at all.

Finding a coefficient in a product of two expansions means pairing up powers that add to the target

To find the coefficient of $x^{k}$ in $(1 + x)^{m} (1 + c x)^{n}$ , expand each bracket (or list its coefficients), then add up every product of a term from the first bracket and a term from the second whose powers of $x$ sum to $k$ .

Worked examples

Find the coefficient of

x^{3}

in the expansion of

(2 + x)^{5}

The general term is $(r 5) (2)^{5 - r} x^{r}$ . For the $x^{3}$ term, set $r = 3$ .
$(3 5) = 10$ and $2^{5 - 3} = 2^{2} = 4$ .
Coefficient $= 10 \times 4 = 40$ .

Expand

(1 - 2 x)^{4}

fully.

General term: $(r 4) (1)^{4 - r} (- 2 x)^{r} = (r 4) (- 2)^{r} x^{r}$ .
For $r = 0, 1, 2, 3, 4$ , $(r 4) (- 2)^{r}$ gives $1, - 8, 24, - 32, 16$ .
So $(1 - 2 x)^{4} = 1 - 8 x + 24 x^{2} - 32 x^{3} + 16 x^{4}$ . Check at $x = 1$ : $1 - 8 + 24 - 32 + 16 = 1$ , which matches $(1 - 2)^{4} = 1$ .

Find the term independent of

x

in the expansion of

(x^{2} + \frac{2}{x})^{6}

General term: $(r 6) (x^{2})^{6 - r} (\frac{2}{x})^{r} = (r 6) 2^{r} x^{2 (6 - r) - r} = (r 6) 2^{r} x^{12 - 3 r}$ .
Independent of $x$ means $12 - 3 r = 0$ , so $r = 4$ .
The term is $(4 6) \times 2^{4} = 15 \times 16 = 240$ .

Find the coefficient of

x^{2}

in the expansion of

(1 + x)^{4} (1 - x)^{3}

$(1 + x)^{4} = 1 + 4 x + 6 x^{2} + 4 x^{3} + x^{4}$ and $(1 - x)^{3} = 1 - 3 x + 3 x^{2} - x^{3}$ .
The $x^{2}$ coefficient in the product is the sum of pairs of terms whose powers add to 2: $(1) (3) + (4) (- 3) + (6) (1)$ .
$= 3 - 12 + 6 = - 3$ .

Mind map

Binomial Expansion

nCr coefficients

nCr = n! / (r!(n-r)!) · nC0 = nCn = 1 · symmetric: nCr = nC(n-r)

General term

T(r+1) = nCr a^(n-r) b^r · (a+b)^n has n+1 terms · r runs from 0 to n

Signs

(a-b)^n: term carries (-1)^r · alternates: + for even r, - for odd r

Finding one term

match power of x to solve for r · independent of x: total power = 0

Products of two expansions

expand each bracket separately · pair up powers that sum to the target

Practise this topic →

Binomial Expansion

Learn it step by step

\(\binom{n}{r}\) counts a choice, and it's the coefficient engine of the expansion

The general term formula finds any term without expanding everything

The coefficients are symmetric: \(\binom{n}{0} = \binom{n}{n} = 1\)

A negative second term makes the signs alternate

To find a specific coefficient, match the power of \(x\) to the right \(r\), then substitute

A term "independent of \(x\)" is the one whose total power of \(x\) is zero

Finding a coefficient in a product of two expansions means pairing up powers that add to the target

Worked examples

Mind map