7.1 Q-21 | Mathematical Induction And Binomial Theorem | Mathematics 11

Exercise 7.1 — Question 21

Expand using the Binomial Theorem and simplify:

$(x - \frac{1}{x})^{6}$

The Binomial Theorem states that for any positive integer $n$ :

$(a + b)^{n} = \sum_{r = 0}^{n} (r n) a^{n - r} b^{r}$

where $(r n) = \frac{n !}{r ! ( n - r )!}$ are the binomial coefficients.

Here $a = x$ , $b = - \frac{1}{x}$ , and $n = 6$ .

Write out each term using $T_{r + 1} = (r 6) x^{6 - r} (- \frac{1}{x})^{r}$ :

$r$	$(r 6)$	$x^{6 - r}$	$(- \frac{1}{x})^{r}$	Term
0	1	$x^{6}$	$1$	$x^{6}$
1	6	$x^{5}$	$- \frac{1}{x}$	$- 6 x^{4}$
2	15	$x^{4}$	$\frac{1}{x ^{2}}$	$15 x^{2}$
3	20	$x^{3}$	$- \frac{1}{x ^{3}}$	$- 20$
4	15	$x^{2}$	$\frac{1}{x ^{4}}$	$\frac{15}{x ^{2}}$
5	6	$x$	$- \frac{1}{x ^{5}}$	$- \frac{6}{x ^{4}}$
6	1	$1$	$\frac{1}{x ^{6}}$	$\frac{1}{x ^{6}}$

$(x - \frac{1}{x})^{6} = x^{6} - 6 x^{4} + 15 x^{2} - 20 + \frac{15}{x ^{2}} - \frac{6}{x ^{4}} + \frac{1}{x ^{6}}$

The expansion has $n + 1 = 7$ terms.
The signs alternate because $b = - \frac{1}{x}$ is negative.
The middle term (4th term, $r = 3$ ) is the constant term $- 20$ , independent of $x$ .
Binomial coefficients are symmetric: $(0 6) = (6 6) = 1$ , $(1 6) = (5 6) = 6$ , $(2 6) = (4 6) = 15$ .