Chapter 12: Q. 12.13 (page 1168)

Write the first five terms of the sequence whose general term is $a_{n} = \frac{2}{n!} .$

Short Answer

Expert verified

The first five terms of the given sequence are $2, 1, \frac{1}{3}, \frac{1}{12} a n d \frac{1}{60} .$

Step by step solution

Step 1. Given information

$\begin{array}{l} Thegivengeneraltermis \\ a_{n} = \frac{2}{n!} \\ Wehavetofindthefirstfivetermofthesequence. \end{array}$

Step 2. Calculating the first five terms

$\begin{array}{l} Weknow \\ If n isapositiveinteger,then \\ \Rightarrow n! = n (n - 1) (n - 2) \dots \end{array}$

We substitute the values 1, 2, 3, 4, 5 into the formula $a_{n} = \frac{2}{n!}$ localid="1645104944104" $\begin{matrix} n = 1 & n = 2 & n = 3 & n = 4 & n = 5 \\ \begin{array}{l} \Rightarrow a_{1} = \frac{2}{1!} \\ = \frac{2}{1} \\ = 2 \end{array} & \begin{array}{l} \Rightarrow a_{2} = \frac{2}{2!} \\ = \frac{2}{2 \cdot 1} \\ = 1 \end{array} & \begin{array}{l} \Rightarrow a_{3} = \frac{2}{3!} \\ = \frac{2}{3 \cdot 2 \cdot 1} \\ = \frac{1}{3} \end{array} & \begin{array}{l} \Rightarrow a_{4} = \frac{2}{4!} \\ = \frac{2}{4 \cdot 3 \cdot 2 \cdot 1} \\ = \frac{1}{12} \end{array} & \begin{array}{l} \Rightarrow a_{5} = \frac{2}{5!} \\ = \frac{2}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \\ = \frac{1}{60} \end{array} \end{matrix}$

Therefore, the first five terms will be $2, 1, \frac{1}{3}, \frac{1}{12} a n d \frac{1}{60} .$

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Write the first five terms of the sequence whose general term is $a_{n} = \frac{2}{n!} .$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Calculating the first five terms

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Write the first five terms of the sequence whose general term isan=2n!.

Short Answer

Step by step solution

Step 1. Given information

Step 2. Calculating the first five terms

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

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Write the first five terms of the sequence whose general term is $a_{n} = \frac{2}{n!} .$