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Q 44.

Expert-verifiedFound in: Page 809

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 $-2,-4,-9,-23,\mathrm{and}-64$.

The given recursively defined sequence is:

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

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

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

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

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 $-2,-4,-9,-23,-64$.

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