Question

b8-taylor and maclaurin series: problem 8
(2 points)
differentiate the maclaurin series for $\frac{1}{1 - 20x}$ twice to find the maclaurin series of $\frac{1}{(1 - 20x)^3}$. index the series so that the 0th term is nonzero.
$\frac{1}{(1 - 20x)^3}=\sum_{n = 0}^{\infty}$
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Explanation:

Step1: Recall the Mac - laurin series of \(\frac{1}{1 - t}\)

The Mac - laurin series of \(\frac{1}{1 - t}=\sum_{n=0}^{\infty}t^{n}\), for \(|t|\lt1\). Let \(t = 20x\), then \(\frac{1}{1-20x}=\sum_{n = 0}^{\infty}(20x)^{n}=\sum_{n=0}^{\infty}20^{n}x^{n}\), for \(|20x|\lt1\) (i.e., \(|x|\lt\frac{1}{20}\)).

Step2: Differentiate \(\frac{1}{1 - 20x}\) once

Differentiate the series term - by - term. Using the formula \(\frac{d}{dx}\sum_{n = 0}^{\infty}a_{n}x^{n}=\sum_{n = 1}^{\infty}na_{n}x^{n - 1}\).
If \(y=\frac{1}{1 - 20x}=\sum_{n=0}^{\infty}20^{n}x^{n}\), then \(y^\prime=\frac{20}{(1 - 20x)^{2}}\).
Differentiating term - by - term: \(\frac{d}{dx}\sum_{n=0}^{\infty}20^{n}x^{n}=\sum_{n = 1}^{\infty}n20^{n}x^{n - 1}\). Let \(m=n - 1\), then \(n=m + 1\) and the series becomes \(\sum_{m=0}^{\infty}(m + 1)20^{m+1}x^{m}\).

Step3: Differentiate \(\frac{20}{(1 - 20x)^{2}}\)

Differentiate \(y^\prime=\frac{20}{(1 - 20x)^{2}}=\sum_{n=0}^{\infty}(n + 1)20^{n+1}x^{n}\) with respect to \(x\) again.
Using the formula \(\frac{d}{dx}\sum_{n = 0}^{\infty}a_{n}x^{n}=\sum_{n = 1}^{\infty}na_{n}x^{n - 1}\).
\(y^{\prime\prime}=\frac{20\times2\times20}{(1 - 20x)^{3}}\).
Differentiating \(\sum_{n=0}^{\infty}(n + 1)20^{n+1}x^{n}\) term - by - term: \(\frac{d}{dx}\sum_{n=0}^{\infty}(n + 1)20^{n+1}x^{n}=\sum_{n = 1}^{\infty}n(n + 1)20^{n+1}x^{n - 1}\). Let \(k=n - 1\), then \(n=k + 1\).
The series becomes \(\sum_{k=0}^{\infty}(k + 1)(k + 2)20^{k+2}x^{k}\).
Since the index of summation is a dummy variable, we can replace \(k\) with \(n\).
So \(\frac{1}{(1 - 20x)^{3}}=\sum_{n = 0}^{\infty}(n + 1)(n + 2)20^{n+2}x^{n}\) for \(|x|\lt\frac{1}{20}\).

Answer:

\(\sum_{n = 0}^{\infty}(n + 1)(n + 2)20^{n+2}x^{n}\)