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Chapter 10: Problem 41

Write an expression for the \(n\) th term of the sequence. (There is more than one correct answer.) $$ 2,-1, \frac{1}{2},-\frac{1}{4}, \frac{1}{8}, \ldots $$

Short Answer

Expert verified
The nth term of the sequence \( 2, -1, 1/2, -1/4, 1/8, \ldots \) can be expressed as \( a_n = 2*(-1)^{n-1}*2^{-n+1} \).

Step by step solution

01

Identify the pattern

The first step is to recognize the pattern if there is any. For the given sequence, we can see that from the second term each term is found by multiplying the preceding term by -1/2.
02

Write down the nth term using the pattern

Once the pattern in the sequence is identified, the next step is write out an expression that can represent the nth term of the sequence. For the given sequence, the nth term can be expressed in terms of \(n\) as: \( a_n = 2 * (-1/2)^{n-1} \).
03

Simplify the nth term

The third step is to simplify this expression to its simplest form if necessary. In this case, the expression for \(a_n\) can be simplified further to: \( a_n = 2*(-1)^{n-1}*2^{-n+1}\).
04

Test the nth term expression

The final step is to verify the expression by testing for some values of the \(n\). For instance, if \(n=1\), \(a_1=2\) and if \(n=2\), \(a_2=-1\). This confirms that our expression is correct as it matches the given sequence.

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Most popular questions from this chapter

Profit The annual revenues for eBay from 2001 through 2006 can be approximated by the model \(a_{n}=540.7 e^{0.42 n}, \quad n=1,2,3,4,5,6\) where \(a_{n}\) is the annual revenue (in millions of dollars) and \(n\) is the year, with \(n=1\) corresponding to \(2001 .\) Use the formula for the sum of a geometric series to approximate the total revenue earned during this 6 -year period. (Source: \(e B a y, I n c .)\)

Verify that the infinite series diverges. $$ \sum_{n=1}^{\infty} \frac{n}{n+1}=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+\cdots $$

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The random variable \(n\) represents the number of units of a product sold per day in a store. The probability distribution of \(n\) is given by \(P(n)\). Find the probability that two units are sold in a given day \([P(2)]\) and show that \(P(0)+P(1)+P(2)+P(3)+\cdots=1\) $$ P(n)=\frac{1}{3}\left(\frac{2}{3}\right)^{n} $$

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