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Chapter 7: Problem 38

Write the series using summation notation (starting with \(k=1\) ). Each series is either an arithmetic series or a geometric series. $$ \frac{7}{16}+\frac{7}{32}+\frac{7}{64}+\cdots+\frac{7}{2^{25}} $$

Short Answer

Expert verified
The given series can be expressed in summation notation as follows: \[ \sum_{k=1}^{22} \frac{7}{16} \cdot \left(\frac{1}{2}\right)^{k-1}. \]

Step by step solution

01

Identify the first term, the common ratio, and the number of terms

The series we are given is: \[ \frac{7}{16}+\frac{7}{32}+\frac{7}{64}+\cdots+\frac{7}{2^{25}} \] First term, \(a_1\), is \(\frac{7}{16}\). Common ratio, \(r\), can be found by dividing the second term by the first term: \[ r = \frac{\frac{7}{32}}{\frac{7}{16}}= \frac{1}{2} \] To find the number of terms, we can use the general formula of the terms of a geometric series, which is: \[ a_n=a_1 \cdot r^{n-1}, \] where \(a_n\) denotes the nth term. In this case, the last term (\(a_n\)) is given as \(\frac{7}{2^{25}}\), so we'll substitute the last term, first term, and common ratio into the general formula and solve for n: \[ \frac{7}{2^{25}} = \frac{7}{16} \cdot \left(\frac{1}{2}\right)^{n-1} \]
02

Solve for the number of terms (n)

To find the value of n, we'll solve the equation from step 1: \[ \frac{7}{2^{25}} = \frac{7}{16} \cdot \left(\frac{1}{2}\right)^{n-1} \] First, we can divide both sides by 7: \[ \frac{1}{2^{25}} = \frac{1}{16} \cdot \left(\frac{1}{2}\right)^{n-1} \] Next, we can express 16 as \(2^4\): \[ \frac{1}{2^{25}} = \frac{1}{2^4} \cdot \left(\frac{1}{2}\right)^{n-1} \] Now, we can multiply both sides by \(2^4\): \[ 2^4\cdot\frac{1}{2^{25}} = \left(\frac{1}{2}\right)^{n-1} \] Simplifying gives: \[ \frac{1}{2^{21}} = \left(\frac{1}{2}\right)^{n-1} \] Now we can conclude that \(n-1 = 21\) since the bases are equal. Thus, we have \(n = 22\).
03

Express the series in summation notation

Now that we have the first term (\(\frac{7}{16}\)), the common ratio (\(\frac{1}{2}\)), and the number of terms (22), we can express the given series in summation notation: \[ \sum_{k=1}^{22} \frac{7}{16} \cdot \left(\frac{1}{2}\right)^{k-1} \] So, the summation notation for the given series is: \[ \sum_{k=1}^{22} \frac{7}{16} \cdot \left(\frac{1}{2}\right)^{k-1} \]

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Most popular questions from this chapter

For Example 2 , the author wanted to find a polynomial \(p\) such that $$ p(1)=1, p(2)=4, p(3)=9, p(4)=16, p(5)=31 $$ Carry out the following steps to see how that polynomial was found. (a) Note that the polynomial $$ (x-2)(x-3)(x-4)(x-5) $$ is 0 for \(x=2,3,4,5\) but is not zero for \(x=1\). By dividing the polynomial above by a suitable number, find a polynomial \(p_{1}\) such that \(p_{1}(1)=1\) and $$ p_{1}(2)=p_{1}(3)=p_{1}(4)=p_{1}(5)=0 $$. (b) Similarly, find a polynomial \(p_{2}\) of degree 4 such that \(p_{2}(2)=1\) and $$ p_{2}(1)=p_{2}(3)=p_{2}(4)=p_{2}(5)=0 $$ (c) Similarly, find polynomials \(p_{j},\) for \(j=3,4,5,\) such that each \(p_{j}\) satisfies \(p_{j}(j)=1\) and \(p_{j}(k)=0\) for values of \(k\) in \\{1,2,3,4,5\\} other than \(j\). (d) Explain why the polynomial \(p\) defined by $$ p=p_{1}+4 p_{2}+9 p_{3}+16 p_{4}+31 p_{5} $$. satisfies $$ p(1)=1, p(2)=4, p(3)=9, p(4)=16, p(5)=31 $$.

Evaluate the geometric series. $$ 1+2+4+\cdots+2^{100} $$

Evaluate \(\lim _{n \rightarrow \infty} \frac{7 n^{2}-4 n+3}{3 n^{2}+5 n+9}\).

Suppose you started an exercise program by riding your bicycle 10 miles on the first day and then you increased the distance you rode by 0.25 miles each day. How many total miles did you ride after 70 days?

Evaluate \(\lim _{n \rightarrow \infty}\left(1+\frac{3}{n}\right)^{n}\).
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