Binomial theorem

Contents

Definitions

Factorial is the number of ways to rearrange nn items in a line.

n!=n(n−1)(n−2)...2(1)n! = n(n-1)(n-2) ... 2(1)

with 0!=1!=1{0! = 1! = 1}

The binomial coefficient is the number of ways to make a group of kk distinct items from nn options.

(nk)=n!r!(n−r)!\displaystyle \binom nk = \frac{n!}{r!(n-r)!}

The binomial expansion formula for n∈Nn \in \mathbb{N}.

(a+b)n=∑r=0n(nr)an−rbr=an+nan−1b+(n2)an−2b2…\begin{align*} (a + b)^n &= \sum_{r=0}^n \binom nr a^{n-r}b^r \\ &= a^n + na^{n-1}b + \binom n2 a^{n-2}b^2 \dots \end{align*}

In the expansion, each term will multiply aa from n−rn - r out of nn factors, and bb from rr factors. The binomial coefficient counts how many times it happens. For example, (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2 with 2ab2ab because it can be aa in first factor times bb in second factor, or bb in first times aa in second.

For HL candidates, we use combination because order does not matter in a⋅b{a \cdot b} and b⋅a{b\cdot a}.

The formula appears again in binomial distribution.

Identities

This means that for every way to choose rr items from nn options, there is a corresponding way to choose n−rn - r items from nn options.

(nr)=(nn−r)\binom{n}r = \binom{n}{n-r}

This row addition can be used to generate Pascal’s triangle.

(nr)+(nr+1)=(n+1r+1)\binom{n}r + \binom{n}{r+1} = \binom{n+1}{r+1}

This means an item has two states: either in a particular group or not in the group. It sums up a row of Pascal’s triangle.

∑r=0n(nr)=2n\sum_{r=0}^n\binom{n}r = 2^{n}

Note that there are n+1n+1 terms in the nnth row of Pascal’s triangle.

Example: Find the x5x^5 term in the expansion of (2x2−13x)4(3x−1)6{\displaystyle\left(2x^2 - \frac{1}{3x}\right)^4(3x - 1)^6}.

If doing this by hand, first generate the Pascal’s triangle up to n=6n = 6.

1             1  1           1  2  1         1  3  3  1       1  4  6  4  1     1  5  10  10  5  1   1  6  15  20  15  6  1\begin{align*} 1 \; & \; & \; & \; & \; & \; & \,\\ 1 \; & 1 \; & \; & \; & \; & \; & \,\\ 1 \; & 2 \; & 1 \; & \; & \; & \; & \,\\ 1 \; & 3 \; & 3 \; & 1 \; & \; & \; & \,\\ 1 \; & 4 \; & 6 \; & 4 \; & 1 \; & \; & \,\\ 1 \; & 5 \; & 10 \; & 10 \; & 5 \; & 1 \; & \, \\ 1 \; & 6 \; & 15 \; & 20 \; & 15 \; & 6 \; & 1 \\ \end{align*}

Applying the binomial expansion formula, the general term for the power of xx of the first binomial in the expanded form is x2(4−m)−m=x8−3mx^{2(4 - m) - m} = x^{8 - 3m}. Similarly, the general term for the second binomial is x6−nx^{6 - n}.

We are looking for

x8−3m⋅x6−n=x58−3m+6−n=59=3m+n\begin{align*}x^{8 - 3m} \cdot x^{6-n} &= x^5\\ 8 - 3m + 6 - n &= 5 \\ 9 &= 3m + n \end{align*}

Solving over natural numbers with m≤4m \leq 4 and n≤6n \leq 6, the first solution is (m,n)=(1,6)(m, n) = (1, 6) and last solution is (m,n)=(3,0)(m, n) = (3, 0). This means (m,n)=(2,3)(m, n) = (2, 3) is also a solution. Note that (0,9)(0, 9) does not work as 9>69 > 6.

Then add up the products of these three multiplications.

We now find the coefficient, ignoring the powers of xx.

C=+(41)24−1(−13)1⋅(66)36−6(−1)6+(42)24−2(−13)2⋅(63)36−3(−1)3+(43)24−3(−13)3⋅(60)36−0(−1)0=+4(2)3(−1)3−1⋅1(3)0(1)+6(2)2(3)−2⋅20(3)3(−1)+4(2)1(−1)(3)−3⋅1(3)6(1)=−323−1440−216=−1666−23=−50003\begin{align*}C = & +\binom412^{4-1}\left(-\frac13\right)^1\cdot\binom663^{6-6}(-1)^6 \\ & + \binom422^{4-2}\left(-\frac13\right)^2\cdot\binom633^{6-3}(-1)^3 \\ & + \binom432^{4-3}\left(-\frac13\right)^3\cdot\binom603^{6-0}(-1)^0 \\ = & +4(2)^3(-1)3^{-1}\cdot 1(3)^0(1) \\&+ 6(2)^2(3)^{-2}\cdot 20(3)^{3} (-1) \\&+ 4(2)^1(-1)(3)^{-3}\cdot 1(3)^6(1) \\ = & -\frac{32}{3} -1440 - 216 \\ = & -1666 - \frac{2}{3} \\ = & -\frac{5000}{3}\end{align*}

The x5x^5 term is −5000x53■\displaystyle -\frac{5000x^5}{3} \qed

Cases

When practising, be sure to cover all of the following cases

  1. given all coefficients in the binomial and nn
  • finding a term in a single binomial
  • finding a term in a product of 2 binomials. This involve listing out all combinations that result in the desired term.
  1. missing a parameter
  • solve for the parameter using given coefficient of the expansion
  1. solving missing nn using graphing calculator; see example on TI-84 Plus
  2. working with terms containing negative exponents
  3. finding binomial coefficients by hand (via Pascal’s triangle) and on calculator
  4. distinguishing between coefficient of the xkx^k term and the xkx^k term

HL: rational exponents

(1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+…(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 +\dots

The formula converges when ∣x∣<1|x|<1. As such, when using it on (a+b)n(a + b)^n, use the correct version below:

if ∣a∣>∣b∣,    (a+b)n=an(1+ba)n\text{if }\lvert a\rvert > \lvert b\rvert, \;\; (a+b)^n = a^n\left(1 + \frac ba\right)^n
if ∣a∣<∣b∣,    (a+b)n=bn(1+ab)n\text{if }\lvert a\rvert < \lvert b\rvert, \;\; (a+b)^n = b^n\left(1 + \frac ab\right)^n

This is also known as the generalized binomial theorem.

Bonus

Check out Bernoulli’s inequality  and a proof using induction.