Question

Find the​ series' interval of convergence​ and, within this​ interval, the sum of the series as a function of x. sum_{n=0}^inftyfrac{(x-4)^{2n}}{36^n}

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asked 2021-01-05
Find the​ series' interval of convergence​ and, within this​ interval, the sum of the series as a function of x.
\(\sum_{n=0}^\infty\frac{(x-4)^{2n}}{36^n}\)

Answers (1)

2021-01-06
Given series is
\(\sum_{n=0}^\infty\frac{(x-4)^{2n}}{36^n}\)
Now,
let \((x-4)^2=y\)
So, the series becomes
\(\sum_{n=0}^\infty\frac{y^n}{36^n}\)
So, \(a_n=\frac{1}{36^n}\)
Then,
\((a_n)^\frac{1}{n}=\frac{1}{36}\)
and
\(\lim_{n\to\infty}(a_n)^{\frac{1}{n}}=\frac{1}{36}\)
So, radius of convergence is 36
Then, the radius of convergence of the given series is 6
Interval of convergence \(=(-6-4,-6+4)=(-10,-2)\)
Now,
\(a_0=1\)
\(a_1=\frac{y}{36}\)
\(a_2=\frac{y^2}{36^2}\)
\(\vdots\)
\(a_n=\frac{y^n}{36^n}\)
So, \(S_n=a_0+a_1+...+a_n\)
\(=1+\frac{y}{36}+\frac{y^2}{36^2}+...+\frac{y^n}{36^n}\)
\(=\frac{1[1-(\frac{y}{36})^n]}{1-\frac{y}{36}}\)
\(=\frac{36[1-\frac{(x-4)^{2n}}{36^n}]}{36-(x-4)^2}\)
\(=\frac{36^n-(x-4)^{2n}}{36^{n-1}[36-(x-4)^2]}\)
Hence, the sum of the series as a function of x is
\(=\frac{36^n-(x-4)^{2n}}{36^{n-1}[36-(x-4)^2]}\)
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