Suggested languages for you:

Americas

Europe

Q 44.

Expert-verified
Found in: Page 809

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# In Problems 37–50, a sequence is defined recursively. Write down the first five terms.${a}_{1}=-2;{a}_{n}=n+3{a}_{n-1}$

The first five terms of the recursively defined sequence are $-2,-4,-9,-23,\mathrm{and}-64$.

See the step by step solution

## Step 1. Write the given information.

The given recursively defined sequence is:

${a}_{1}=-2\phantom{\rule{0ex}{0ex}}{a}_{n}=n+3{a}_{n-1}$

## Step 2. Determine the first term and find out the second term.

The first term is $\left({a}_{1}\right)=3$.

Now substitute 2 for n in the given formula ${a}_{n}=n+3{a}_{n-1}$ to get the second term,

role="math" localid="1646730495162" ${a}_{2}=2+3{a}_{2-1}\phantom{\rule{0ex}{0ex}}=2+3{a}_{1}\phantom{\rule{0ex}{0ex}}=2+3\left(-2\right)\phantom{\rule{0ex}{0ex}}=2-6\phantom{\rule{0ex}{0ex}}=-4$

## Step 3. Find the 3rd and 4th terms.

Similarly substitute 3, 4 for n in the given formula to get the third and fourth term,

${a}_{3}=3+3{a}_{3-1}\phantom{\rule{0ex}{0ex}}=3+3{a}_{2}\phantom{\rule{0ex}{0ex}}=3+3\left(-4\right)\phantom{\rule{0ex}{0ex}}=-9\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{a}_{4}=4+3{a}_{4-1}\phantom{\rule{0ex}{0ex}}=4+3{a}_{3}\phantom{\rule{0ex}{0ex}}=4+3\left(-9\right)\phantom{\rule{0ex}{0ex}}=4-27\phantom{\rule{0ex}{0ex}}=-23$

## Step 4. Find the 5th term.

Now substitute 5 for n in the given formula to get the 5th term,

${a}_{5}=5+3{a}_{5-1}\phantom{\rule{0ex}{0ex}}=5+3{a}_{4}\phantom{\rule{0ex}{0ex}}=5+3\left(-23\right)\phantom{\rule{0ex}{0ex}}=5-69\phantom{\rule{0ex}{0ex}}=-64$

Therefore, the first five terms are $-2,-4,-9,-23,-64$.