Question

Use the binomial series to expand the function as a power series. 3 (6 + x)3 Σ(1 ) 00 n=0 State the radius of convergence, R.

Answer 1

The radius of convergence
`\(R\)` for the original series
`\((3)/((6+x)^3)\)` is
`\(6\)`.

The power series expansion of the function
`\((3)/((6+x)^3)\)` up to six terms is given by:

`\[(1)/(72) - (x)/(144) + (x^2)/(432) - (5x^3)/(7776) + (5x^4)/(31104) - (7x^5)/(186624) + \cdots\]`

Now, let's find the radius of convergence
`\(R\)` for this series.

The general term of the binomial series
`\((1)/((1-x)^n)\)` is given by
`\(\binom{n+k-1}{k}x^k\)`, where
`\(\binom{n}{k}\)` is the binomial coefficient. In our case, for
`\((3)/((6+x)^3)\)`, we can rewrite the function as
`\((3)/(6^3) \cdot (1)/((1+(x)/(6))^3)\)` and apply the binomial series expansion with
`\(n=3\)` and
`\(a=(1)/(6)\)`.

The radius of convergence of the binomial series
`\((1)/((1-x)^n)\)` is 1, so for our transformed series, the radius of convergence will be
`\(|a| \cdot R = (1)/(6) \cdot 1 = (1)/(6)\)`.

Therefore, the answer is
`\(6\)`.

The complete question is here:

Use the binomial series to expand the function as a power series.

`$$\begin{aligned}& (3)/((6+x)^3) \\& \sum_(n=0)^(\infty)(\perp)\end{aligned}$$`(______)

State the radius of convergence,
`$R$`.