The binomial theorem

You know $(a+b)^2=a^2+2ab+b^2$. What about $(a+b)^5$? Multiplying five brackets by hand is slow and error-prone, but there is a pattern, and it comes from counting: each term arises by choosing $a$ or $b$ from each bracket. That is exactly what the binomial coefficients $\binom nk$ count.

Small powers and Pascal's triangle

$$\begin{aligned}(a+b)^2&=a^2+2ab+b^2\\(a+b)^3&=a^3+3a^2b+3ab^2+b^3\\(a+b)^4&=a^4+4a^3b+6a^2b^2+4ab^3+b^4\end{aligned}$$

The coefficients $1, 2, 1$ then $1, 3, 3, 1$ then $1, 4, 6, 4, 1$ form Pascal's triangle: each number is the sum of the two above it, because $\binom nk+\binom n{k+1}=\binom{n+1}{k+1}$.

$$\begin{array}{ccccccccccccc} & & & & & & 1 & & & & & & \\ & & & & & 1 & & 1 & & & & & \\ & & & & 1 & & 2 & & 1 & & & & \\ & & & 1 & & 3 & & 3 & & 1 & & & \\ & & 1 & & 4 & & 6 & & 4 & & 1 & & \\ & 1 & & 5 & & 10 & & 10 & & 5 & & 1 & \\ 1 & & 6 & & 15 & & 20 & & 15 & & 6 & & 1 \end{array}$$

Row $n$ of the triangle lists $\binom n0, \binom n1, \dots, \binom nn$.

The binomial theorem

One coefficient only

You rarely need the whole expansion. Write the general term and pick the right $k$.

Sum of coefficients and approximations

Putting $x=1$ in a polynomial gives the sum of its coefficients. Putting $a=b=1$ in the theorem gives $\sum_{k}\binom nk=2^n$.

Key terms

Summary

• $(a+b)^n=\sum_{k=0}^n\binom nka^{n-k}b^k$, with $n+1$ terms.
• The coefficients are row $n$ of Pascal's triangle.
• $T_{k+1}=\binom nka^{n-k}b^k$; set the power of $x$ to find $k$.
• Sum of coefficients: put $x=1$. Approximation: use the first few terms.

Practice

Show solutions

1. Row $5$: $1, 5, 10, 10, 5, 1$. $(x+1)^5=x^5+5x^4+10x^3+10x^2+5x+1$.
2. Row $4$ with $b=-2$: $x^4-8x^3+24x^2-32x+16$.
3. $T_{k+1}=\binom5k3^{5-k}x^k$. For $k=2$: $\binom52\cdot3^3=10\cdot27=270$.
4. $T_5=T_{4+1}=\binom74a^3b^4=35\,a^3b^4$.
5. $T_{k+1}=\binom8k(-x)^k$. For $k=4$: $\binom84=70$ (the sign is $+$ since $(-1)^4=1$).
6. Put $x=1$: $(2-3)^4=1$.
7. This is $(1+1)^6=2^6=64$.
8. $T_{k+1}=\binom6k(x^2)^{6-k}x^{-k}=\binom6kx^{12-3k}$. The power is $0$ when $k=4$: $\binom64=15$.
9. $1.02^6\approx1+6(0.02)+15(0.02)^2=1+0.12+0.006=1.126$. (Exact value $1.1262$ to four decimals.)
10. The coefficient of $x^3$ is $\binom53+\binom63=10+20=30$.