(i) Number of terms in the expansion of $\left(x - \frac{1}{x}\right)^4$ (1)
(ii) Write the expansion of $\left(x - \frac{1}{x}\right)^4$ (3)

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> [!TLDR]
> The binomial expansion of $(a+b)^n$ always has $n+1$ terms; apply the binomial theorem term by term with $a=x$, $b=-\frac{1}{x}$.
>
> **(i)** For $(x - \frac{1}{x})^4$, $n=4$, so the number of terms is $n+1 = 5$.
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> **(ii)** By the binomial theorem, $(x-\frac1x)^4 = \displaystyle\sum_{k=0}^{4} \binom{4}{k}x^{4-k}\left(-\frac1x\right)^k = \sum_{k=0}^4 \binom{4}{k}(-1)^k x^{4-2k}$
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> $$k=0:\ \binom{4}{0}x^4 = x^4$$
> $$k=1:\ -\binom{4}{1}x^2 = -4x^2$$
> $$k=2:\ \binom{4}{2}x^0 = 6$$
> $$k=3:\ -\binom{4}{3}x^{-2} = -\frac{4}{x^2}$$ …

Power	Expansion
$(x + y)^0$	$1$
$(x + y)^1$	$x + y$
$(x + y)^2$	$x^2 + 2xy + y^2$
$(x + y)^3$	$x^3 + 3x^2y + 3xy^2 + y^3$
$(x + y)^4$	$x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$

$k$	$\binom{4}{k}$	$x^{4-k}$	$y^k$	Term
0	$\binom{4}{0}=1$	$x^4$	$y^0$	$1 \cdot x^4$
1	$\binom{4}{1}=4$	$x^3$	$y^1$	$4x^3y$
2	$\binom{4}{2}=6$	$x^2$	$y^2$	$6x^2y^2$
3	$\binom{4}{3}=4$	$x^1$	$y^3$	$4xy^3$
4	$\binom{4}{4}=1$	$x^0$	$y^4$	$1 \cdot y^4$

Q.(i) Number of terms in the expansion of $\left(x - \frac{1}{x}\right)^4$

Concept understanding — Binomial Theorem Expansion

The Problem That Started It All

The Pattern You Already Know

The Precise Statement

Where Do Those Coefficients Come From?

The General Term

What If It's Not Just $x$ and $y$ ? …

Q.(i) Number of terms in the expansion of (x−1x)4\left(x - \frac{1}{x}\right)^4(x−x1​)4

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Q.(i) Number of terms in the expansion of $\left(x - \frac{1}{x}\right)^4$

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