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Magazine Chapter 11: Problem 29
Find the first five terms of the infinite sequence whose nth term is given. $$a_{n}=\frac{2^{n}}{n !}$$
Short Answer
Expert verified
The first five terms are: 2, 2, \(\frac{4}{3}\), \(\frac{2}{3}\), \(\frac{4}{15}\)
Step by step solution
01
- Substitute n = 1
The given term formula is \(a_{n} = \frac{2^{n}}{n!}\). Start by substituting \(n = 1\): \[a_{1} = \frac{2^{1}}{1!}\]
02
- Calculate the first term
Solve for \(a_{1}\): \[a_{1} = \frac{2}{1} = 2\]
03
- Substitute n = 2
Now, substitute \(n = 2\): \[a_{2} = \frac{2^{2}}{2!}\]
04
- Calculate the second term
Solve for \(a_{2}\): \[a_{2} = \frac{4}{2} = 2\]
05
- Substitute n = 3
Now, substitute \(n = 3\): \[a_{3} = \frac{2^{3}}{3!}\]
06
- Calculate the third term
Solve for \(a_{3}\): \[a_{3} = \frac{8}{6} = \frac{4}{3}\]
07
- Substitute n = 4
Now, substitute \(n = 4\): \[a_{4} = \frac{2^{4}}{4!}\]
08
- Calculate the fourth term
Solve for \(a_{4}\): \[a_{4} = \frac{16}{24} = \frac{2}{3}\]
09
- Substitute n = 5
Finally, substitute \(n = 5\): \[a_{5} = \frac{2^{5}}{5!}\]
10
- Calculate the fifth term
Solve for \(a_{5}\): \[a_{5} = \frac{32}{120} = \frac{4}{15}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sequence Terms
In mathematics, a sequence is an ordered list of numbers following a specific pattern. The sequence defined by a formula for its nth term allows us to generate subsequent terms. For the given sequence, each term is created using the expression: \[a_{n} = \frac{2^{n}}{n!}\] To find the first five terms, substitute different values of 'n' one by one:
- For n = 1, you get: \[a_{1} = \frac{2^{1}}{1!} = 2\]
- Next, for n = 2, it becomes: \[a_{2} = \frac{2^{2}}{2!} = 2\]
- For n = 3, the term is: \[a_{3} = \frac{2^{3}}{3!} = \frac{4}{3}\]
- Similarly, for n = 4, you find: \[a_{4} = \frac{2^{4}}{4!} = \frac{2}{3}\]
- Finally, for n = 5, the term is: \[a_{5} = \frac{2^{5}}{5!} = \frac{4}{15}\]
Factorials
Factorials are a key part of many mathematical sequences. The factorial of a positive integer 'n' (denoted as \(n!\) ) is the product of all positive integers up to 'n'. For example:
- \(1! = 1\)
- \(2! = 2 \times 1 = 2\)
- \(3! = 3 \times 2 \times 1 = 6\)
- \(4! = 4 \times 3 \times 2 \times 1 = 24\)
- \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
- For n = 1, \(\frac{2^{1}}{1!} = 2\)
- For n = 2, \(\frac{2^{2}}{2!} = 2\)
- For n = 3, \(\frac{2^{3}}{3!} = \frac{4}{3}\)
- For n = 4, \(\frac{2^{4}}{4!} = \frac{2}{3}\)
- For n = 5, \(\frac{2^{5}}{5!} = \frac{4}{15}\)
Substitution Method
The substitution method is straightforward and crucial for understanding and calculating terms in a sequence. You simply replace the variable 'n' in the given formula with specific values. Here, we'll see how substituting manageable numbers helps break down the problem clearly:
- Substituting \(n = 1\) into \(a_{n} = \frac{2^{n}}{n!}\) gives \(a_{1} = \frac{2}{1} = 2\)
- Substituting \(n = 2\) gives \(a_{2} = \frac{4}{2} = 2\)
- For \(n = 3\), we get \(a_{3} = \frac{8}{6} = \frac{4}{3}\)
- Using \(n = 4\), yields \(a_{4} = \frac{16}{24} = \frac{2}{3}\)
- Finally, for \(n = 5\), it results in \(a_{5} = \frac{32}{120} = \frac{4}{15}\)
Series Calculations
When working with infinite sequences, terms calculated via substitution method can be summed to form a series. Series calculations involve adding up all terms of a sequence to see convergence or divergence trends. Here we sum up the first five terms from our example sequence: \[2 + 2 + \frac{4}{3} + \frac{2}{3} + \frac{4}{15}\] To make the series calculation easier, we note each fraction's common denominators:
- Convert \(\frac{4}{3}\) and \(\frac{2}{3}\), common denominator 3:
- \(\frac{4}{3} = \frac{4}{3}\)
- \(\frac{2}{3} = \frac{2}{3}\)
- Next, add these simpler terms for clarity:
- Convert fractions for simpler summation:
- For \(\frac{4}{15}\), common denominator 15 remains:
- \(\frac{4}{15} = \frac{4}{15}\)
