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Chapter 8: Problem 1

Write the first five terms of each geometric sequence. $$a_{1}=5, \quad r=3$$

Short Answer

Expert verified
The first five terms of the geometric sequence are 5, 15, 45, 135, 405.

Step by step solution

01

Identify the first term and the common ratio

In this geometric sequence, the first term, known as \(a_{1}\), is 5 and the common ratio (r) is 3. This ratio is the number we multiply by to get from one term to the next.
02

Compute the second term

To compute the second term of the sequence, we multiply the first term by the common ratio. So, \(a_{2}\) = \(a_{1} \times r\) = 5 * 3 = 15.
03

Compute the third term

We repeat the process for the third term by multiplying the second term by the common ratio. So, \(a_{3}\) = \(a_{2} \times r\) = 15 * 3 = 45.
04

Compute the fourth term

The fourth term is computed by multiplying the third term by the common ratio. So, \(a_{4}\) = \(a_{3} \times r\) = 45 * 3 = 135.
05

Compute the fifth term

Lastly, the fifth term is computed by multiplying the fourth term by the common ratio. Hence, \(a_{5}\) = \(a_{4} \times r\) = 135 * 3 = 405.

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Most popular questions from this chapter

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.

Find the term indicated in each expansion. \((x-1)^{10} ;\) fifth term

If \(f(x)=x^{5},\) find \(\frac{f(x+h)-f(x)}{h}\) and simplify.

Find the term in the expansion of \(\left(x^{2}+y^{2}\right)^{5}\) containing \(x^{4}\) as a factor.

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