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Chapter 14: Problem 29

Find each indicated sum. $$\sum_{i=1}^{5} \frac{i !}{(i-1) !}$$

Short Answer

Expert verified
The sum of the given series is 15.

Step by step solution

01

Understanding Factorials and Summation

Firstly, let's understand what a factorial is. For any positive integer \(n\), \(n!\) means \(n \times (n-1) \times (n-2) \times ... \times 2 \times 1\). Now, let's rewrite the given summation problem, noting that \((i-1)! = i !/i\): \[\sum_{i=1}^{5} \frac{i !}{(i-1) !} = \sum_{i=1}^{5} i\]
02

Calculate The Sum

Now it results in a simple arithmetic series from 1 to 5, which is easy to calculate:\[\sum_{i=1}^{5} i = 1+2+3+4+5\]
03

Final Summation Value

Adding these numbers together, we get:\[1+2+3+4+5 = 15\]

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Most popular questions from this chapter

What is a geometric sequence? Give an example with your explanation.

Factor: \(27 x^{3}-8\) (Section 6.4, Example 8)

Find a general term, \(a_{n},\) for each sequence. More than one answer may be possible. Evaluate without using a calculator: \(\frac{600 !}{599 !}\)

Find a general term, \(a_{n},\) for each sequence. More than one answer may be possible. $$-1,1,-1,1, \dots$$

If \(f(x)=x^{2}+5 x\) and \(g(x)=2 x-3,\) find \(f(g(x))\) and \(g(f(x))\) (Section 8.4, Example 1)
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