Use the binomial series to expand the function as a power series. 3/(6 + x)³

Solution:

Given, the function is 3/(6 + x)³

We have to expand the function using binomial series.

The binomial expansion for (1 + x)ⁿ is given by \(1+nx+\frac{n(n-1)x^{2}}{2!}+\frac{n(n-1)(n-2)x^{3}}{3!}+........\)

The function can be written as 3/6³(1 + (x/6))³

6³ = 216

3/216 = 1/72

So, the function becomes (1/72)(1 + x/6)^-3

Using binomial expansion for n = -3

(1 + x/6)^-3 = \(1+(-3)\frac{x}{6}+\frac{(-3)(-3-1)(\frac{x}{6})^{2}}{2!}+\frac{(-3)(-3-1)(-3-2)(\frac{x}{6})^{3}}{3!}+....\)

\(\\=1-\frac{3x}{6}+\frac{(-3)(-4)(\frac{x}{6})^{2}}{2}+\frac{(-3)(-4)(-5)(\frac{x}{6})^{3}}{3!}+....\\=1-\frac{x}{2}+\frac{12(\frac{x^{2}}{36})}{2}-\frac{60(\frac{x^{3}}{216})}{6}+....\\=1-\frac{x}{2}+(\frac{6x^{2}}{36})-\frac{10x^{3}}{216}+....\\=1-\frac{x}{2}+\frac{x^{2}}{6}-\frac{5x^{3}}{108}+.....\)

Therefore, (1/72)(1+ x/6)^-3 = \(=\frac{1}{72}[1-\frac{x}{2}+\frac{x^{2}}{6}-\frac{5x^{3}}{108}+.....]\).

Use the binomial series to expand the function as a power series. 3/(6 + x)³

Summary:

Using binomial series to expand the function 3/(6 + x)³ as a power series is \(=\frac{1}{72}[1-\frac{x}{2}+\frac{x^{2}}{6}-\frac{5x^{3}}{108}+.....]\).