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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.${a}_{1}=-1;{a}_{2}=1;{a}_{n}={a}_{n-2}+n{a}_{n-1}$

The first five terms of the recursively defined sequence are $-1,1,2,9,\mathrm{and}47$.

See the step by step solution

## Step 1. Write the given information.

The given recursively defined sequence is:

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

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

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

The second term islocalid="1646734892458" $\left({a}_{2}\right)=1$

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

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

## Step 3. Find the 4th and 5th terms.

Similarly substitute 4 and 5 for n in the given formula to get the fourth and fifth term,

${a}_{4}={a}_{4-2}+4{a}_{4-1}\phantom{\rule{0ex}{0ex}}={a}_{2}+4{a}_{3}\phantom{\rule{0ex}{0ex}}=1+4\left(2\right)\phantom{\rule{0ex}{0ex}}=9\phantom{\rule{0ex}{0ex}}{a}_{5}={a}_{5-2}+5{a}_{5-1}\phantom{\rule{0ex}{0ex}}={a}_{3}+5{a}_{4}\phantom{\rule{0ex}{0ex}}=2+5\left(9\right)\phantom{\rule{0ex}{0ex}}=47$

Therefore, the first five terms are $-1,1,2,9,47$. ### Want to see more solutions like these? 