1.9 The binomial theoremIB Maths: Analysis and Approaches HL: Revision notes

For $n\in\mathbb{N}$,$(a+b)^n = a^n + \binom n1a^{n-1}b + \binom n2a^{n-2}b^2 + \cdots + \binom nra^{n-r}b^r + \cdots + b^n.$There are $n+1$ terms. In each term the powers of $a$ and $b$ add to $n$. The general term is $\binom nr a^{n-r}b^r$ (this is in the formula booklet).

The binomial coefficient $\binom nr = {}^nC_r = \frac{n!}{r!(n-r)!}$. For small $n$ read them from Pascal's triangle, where each entry is the sum of the two above it:

row 5: $1, 5, 10, 10, 5, 1$ row 6: $1, 6, 15, 20, 15, 6, 1$

So $\binom73 = 15+20 = 35$. You should be able to find $\binom nr$ with the formula and with your GDC (the nCr function). The triangle is symmetric: $\binom nr = \binom n{n-r}$.

Write the general term, simplify the powers of $x$, and solve for $r$.

Example: the term independent of $x$ in $\left(x-\frac2x\right)^6$. General term $\binom6r x^{6-r}\left(-\frac2x\right)^r = \binom6r(-2)^r x^{6-2r}$. Independent of $x$ means $6-2r = 0$, so $r = 3$: $20\times(-8) = -160$.

For a product such as $(1-x)(2+x)^5$, list every pair of terms whose powers of $x$ add to the one you want: the $x^2$ coefficient is $1\times[x^2] + (-1)\times[x^1] = 80 - 80 = 0$.

If coefficients are given, form equations. From $(1+kx)^n$ with $x$-coefficient 12 and $x^2$-coefficient 60: $nk = 12$ and $\frac{n(n-1)}{2}k^2 = 60$. Substitute $k = \frac{12}{n}$ to get $n = 6$, then $k = 2$.

When $x$ is small, the first few terms of $(1+x)^n$ give a good approximation, because higher powers of $x$ are tiny. $1.02^8 = (1+0.02)^8 \approx 1 + 8(0.02) + 28(0.02)^2 + 56(0.02)^3 = 1.171648$.

If $x>0$ all omitted terms are positive, so the truncated sum is an underestimate.