/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 Find each indicated sum. $$\su... [FREE SOLUTION] | 91Ó°ÊÓ

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Find each indicated sum. $$\sum_{i=2}^{4}\left(-\frac{1}{3}\right)^{i}$$

Short Answer

Expert verified
The sum of the series is -\frac{37}{243}.

Step by step solution

01

Understand the series

In this problem, we are working with a geometric series whose formula is represented as \(-\frac{1}{3}^i\), where \(i\) is the step. This series begins at \(i = 2\) and continues until \(i = 4\)
02

Calculate the terms

Calculate each term in the series individually. To do it, replace \(i\) in the formula with the each individual value from 2 to 4. For \(i=2\), the term is \(-\frac{1}{3}^2 = -\frac{1}{9}\). For \(i=3\), the term is \(-\frac{1}{3}^3 = -\frac{1}{27}\). For \(i=4\), the term is \(-\frac{1}{3}^4 = -\frac{1}{81}\).
03

Sum up all the terms

Finally, sum up all the individual terms together to get the sum of the series. The sum therefore is \(-\frac{1}{9} - \frac{1}{27} - \frac{1}{81} = - \frac{37}{243}\).

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Most popular questions from this chapter

Use the Binomial Theorem to expand each binomial and express the result in simplified form. $$ \left(x^{2}+y\right)^{4} $$

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Follow the outline on the next page to use mathematical induction to prove that $$ \begin{aligned}(a+b)^{n}=\left(\begin{array}{c}n \\\0\end{array}\right) a^{n}+\left(\begin{array}{c}n \\\1\end{array}\right) a^{n-1} b+\left(\begin{array}{c}n \\\2\end{array}\right) a^{n-2} b^{2} \\\\+\cdots+\left(\begin{array}{c}n \\\n-1\end{array}\right) a b^{n-1}+\left(\begin{array}{c}n \\\n\end{array}\right) b^{n}\end{aligned} $$ a. Verify the formula for \(n=1\) b. Replace \(n\) with \(k\) and write the statement that is assumed true. Replace \(n\) with \(k+1\) and write the statement that must be proved. c. Multiply both sides of the statement assumed to be true by \(a+b .\) Add exponents on the left. On the right, distribute \(a\) and \(b,\) respectively. d. Collect like terms on the right. At this point, you should have $$ \begin{aligned}&(a+b)^{k+1}=\left(\begin{array}{l}k \\\0\end{array}\right) a^{k+1}+\left[\left(\begin{array}{l}k \\\0\end{array}\right)+\left(\begin{array}{l}k \\\1\end{array}\right)\right] a^{k} b\\\&\begin{array}{l}+\left[\left(\begin{array}{c}k \\\1\end{array}\right)+\left(\begin{array}{c}k \\\2\end{array}\right)\right] a^{k-1} b^{2}+\left[\left(\begin{array}{c}k \\\2\end{array}\right)+\left(\begin{array}{c}k \\\3\end{array}\right)\right] a^{k-2} b^{3} \\\\+\cdots+\left[\left(\begin{array}{c}k \\\k-1\end{array}\right)+\left(\begin{array}{c}k \\\k\end{array}\right)\right] a b^{k}+\left(\begin{array}{c}k \\\k\end{array}\right) b^{k+1} \end{array}\end{aligned} $$ e. Use the result of Exercise 74 to add the binomial sums in brackets. For example, because \(\left(\begin{array}{l}n \\\ r\end{array}\right)+\left(\begin{array}{c}n \\\ r+1\end{array}\right)$$=\left(\begin{array}{l}n+1 \\ r+1\end{array}\right),\) then \(\left(\begin{array}{l}k \\ 0\end{array}\right)+\left(\begin{array}{l}k \\\ 1\end{array}\right)=\left(\begin{array}{c}k+1 \\\1\end{array}\right)\) and\(\left(\begin{array}{l}k \\ 1\end{array}\right)+\left(\begin{array}{l}k \\\2\end{array}\right)=\left(\begin{array}{c}k+1 \\ 2\end{array}\right)\) f. Because \(\left(\begin{array}{l}k \\\ 0\end{array}\right)=\left(\begin{array}{c}k+1 \\ 0\end{array}\right) \quad\) (why?) and \(\left(\begin{array}{l}k \\ k\end{array}\right)=\) \(\left(\begin{array}{l}k+1 \\ k+1\end{array}\right)\) (why?), substitute these results and the results from part (e) into the equation in part (d). This should give the statement that we were required to prove in the second step of the mathematical induction process.

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