b8-taylor and maclaurin series: problem 8\n(2 points)\ndifferentiate the maclaurin series for $\\frac{1}{1…
b8-taylor and maclaurin series: problem 8\n(2 points)\ndifferentiate 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.\n$\\frac{1}{(1 - 20x)^3}=\\sum_{n = 0}^{\\infty}$\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have 15 attempts remaining.\npage generated october 21, 2025, 10:37:52 pm edt\nwebwork © 1996 - 2025 | theme: math4 | ww_version: 2.20 | pg_version 2.20\nthe webwork project
Answer
Answer:
(\sum_{n = 0}^{\infty}(n + 1)(n + 2)20^{n+2}x^{n})
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}).