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Chapter 9: Problem 103

Express \(\sum_{k=1}^{\infty} \frac{6^{k}}{\left(3^{k+1}-2^{k+1}\right)\left(3^{k}-2^{k}\right)}\) as a rational number.

Short Answer

Expert verified
\(\frac{4}{5}\)

Step by step solution

01

Express Terms

Rewrite the term of the series \(\frac{6^{k}}{\left(3^{k+1}-2^{k+1}\right)\left(3^{k}-2^{k}\right)}\) as a difference of fractions where each denominator has only one variable. This can be written as \(\frac{2}{3^{k}-2^{k}}-\frac{1}{3^{k+1}-2^{k+1}}\). This is done by factoring out \(6^{k} = 2\cdot3^{k}\) from numerator and then splitting up the single fraction into two.
02

Identify Telescoping Series

These series are called telescoping because many of the terms cancel out when the sum is expanded. Now, the sequence can be written as \(S = 2\cdot(\frac{1}{2} - \frac{1}{2^2}) + 2\cdot(\frac{1}{2^2} - \frac{1}{2^3}) + \ldots - \frac{1}{3^{k+1}-2^{k+1}}\).
03

Simplify the Series

Observe that each pair of subsequent fractions will cancel out one of their terms, so that all but two terms cancel out at the end, i.e. \[S = 2(\frac{1}{2} - \frac{1}{3^{2}-2^{2}})\].
04

Compute the Remaining Terms

Calculate the remaining term. The expression simplifies to \[S = 1 - \frac{1}{5} = \frac{4}{5}.\]

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Most popular questions from this chapter

Finding a Maclaurin Series In Exercises 53 and \(54,\) find a Maclaurin series for \(f(x) .\) $$ f(x)=\int_{0}^{x} \sqrt{1+t^{3}} d t $$

Finding a Taylor Series In Exercises \(1-12,\) use the definition of Taylor series to find the Taylor series, centered at \(c,\) for the function. $$ f(x)=\frac{1}{1-x}, \quad c=2 $$

Finding a Limit In Exercises \(59-62,\) use the series representation of the function \(f\) to find \(\lim _{x \rightarrow 0} f(x)\) (if it exists). $$ f(x)=\frac{\sin x}{x} $$

Finding a Taylor Series In Exercises \(1-12,\) use the definition of Taylor series to find the Taylor series, centered at \(c,\) for the function. $$ f(x)=e^{2 x}, \quad c=0 $$

Verifying a Sum In Exercises \(55-58\) , verify the sum. Then use a graphing utility to approximate the sum with an error of less than \(0.0001 .\) $$ \sum_{n=0}^{\infty}(-1)^{n}\left[\frac{1}{(2 n+1) !}\right]=\sin 1 $$
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