# How do you use the binomial series to expand  (x - sqrt 2)^5?

Feb 27, 2016

Binomial Expansion: ${\left(a + b\right)}^{n} = \setminus {\sum}_{k = 0}^{n} C \left(n , k\right) {a}^{n - k} {b}^{k}$, where $C \left(n , k\right)$ is the the binomial coefficient which is defined as:

C(n,k) \equiv \frac{n!}{k!(n-k)!}.

${\left(a - b\right)}^{n} = \setminus {\sum}_{k = 0}^{n} C \left(n , k\right) {a}^{n - k} {\left(- 1. b\right)}^{k}$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus {\sum}_{k = 0}^{n} {\left(- 1\right)}^{k} C \left(n , k\right) {a}^{k} {b}^{n - k}$.

Solution: a=x; \qquad b=\sqrt{2}; \qquad n=5
${\left(x - \setminus \sqrt{2}\right)}^{5} = {\left(- 1\right)}^{0} C \left(5 , 0\right) {x}^{5} + {\left(- 1\right)}^{1} C \left(5 , 1\right) {x}^{4} {\left(\setminus \sqrt{2}\right)}^{3} +$
${\left(- 1\right)}^{2} C \left(5 , 2\right) {x}^{3} {\left(\setminus \sqrt{2}\right)}^{2} + {\left(- 1\right)}^{3} C \left(5 , 3\right) {x}^{2} {\left(\setminus \sqrt{2}\right)}^{3} +$
${\left(- 1\right)}^{4} C \left(5 , 4\right) x {\left(\setminus \sqrt{2}\right)}^{4} + {\left(- 1\right)}^{5} C \left(5 , 5\right) {\left(\setminus \sqrt{2}\right)}^{5}$

Binomial Coefficients: $C \left(n , k\right) = C \left(n , n - k\right)$
C(5,0)=C(5,5)=1; \qquad C(5,1)=C(5,4)=5;
$C \left(5 , 2\right) = C \left(5 , 3\right) = 10$

${\left(x - \setminus \sqrt{2}\right)}^{5} = {x}^{5} - 5 \setminus \sqrt{2} {x}^{4} + 20 {x}^{3} - 20 \setminus \sqrt{2} {x}^{2} + 20 x - 4 \setminus \sqrt{2}$