Chapter 9, Problem 76RE

To determine

Write first five terms of the sequence, calculate first and second diff of sequence also state model of the sequence

Answer to Problem 76RE

First five terms of the sequence are, $–3,–7,–13,–21,–31$

First difference: $–4–6–8–10$

Second difference: $–2–2–2$

The second differences are all same, the sequence has a perfect quadratic model

Explanation of Solution

Given:

Consider,

  $a_1=–3,{\text{ a}}_n{\text{=a}}_{n-1}–2n$

Considering the given,

  $a_1=–3,{\text{ a}}_n{\text{=a}}_{n-1}–2n$

So,

  $\begin{array}{l}{a}_2=a_1–2\left(2\right)\hfill \\ =–3–4\hfill \\ =–7\hfill \end{array}$

  $\begin{array}{l}\begin{array}{l}{a}_3=a_2–2\left(3\right)\hfill \\ =–7–6\hfill \end{array} \\ =–13\end{array}$

  $\begin{array}{l}\begin{array}{l}{a}_4=a_3–2\left(4\right)\hfill \\ =–13–8\hfill \end{array} \\ =–21\end{array}$

  $\begin{array}{l}\begin{array}{l}{a}_5=a_4–2\left(5\right)\hfill \\ =–21–10\hfill \end{array}\\ =–31\end{array}$

Hence first five terms of the sequence are,

  $–3,–7,–13,–21,–31$

The first difference of the sequence are found by subtracting consecutive terms.

The second differences are found by subtracting consecutive first differences.

Thus,

  $\begin{array}{l}n:12345\hfill \\ a_n:–3–7–13–21–31\hfill \end{array}$

First difference: $–4–6–8–10$

Second difference: $–2–2–2$

Since the second differences are all same, the sequence has a perfect quadratic model.