# How do you find the first five terms given a_1=4, a_2=-3, a_(n+2)=a_(n+1)+2a_n?

Nov 9, 2016

The first five terms of the sequence are $4 , - 3 , 5 , - 1$, and $9$.

#### Explanation:

You already have the first two terms. The third term through the fifth term will be found using the given formula.

${a}_{n + 2} = {a}_{n + 1} + 2 {a}_{n}$

Notice that in order to find the value of each of these terms, we have to use the subscript on the left side of the formula to find the value of $n$ for each term.

${a}_{1} = 4$

${a}_{2} = - 3$

For the third term:
$3 = n + 2$
$3 - 2 = n + 2 - 2$
$1 = n$
So,
${a}_{3} = {a}_{1 + 1} + 2 {a}_{1}$
${a}_{3} = {a}_{2} + 2 \left(4\right)$
${a}_{3} = - 3 + 8$
${a}_{3} = 5$

For the fourth term:
$4 = n + 2$
$4 - 2 = n + 2 - 2$
$2 = n$
So,
${a}_{4} = {a}_{2 + 1} + 2 {a}_{2}$
${a}_{4} = {a}_{3} + 2 \left(- 3\right)$
${a}_{4} = 5 + - 6$
${a}_{4} = - 1$

For the fifth term:
$5 = n + 2$
$5 - 2 = n + 2 - 2$
$3 = n$
So,
${a}_{5} = {a}_{3 + 1} + 2 {a}_{3}$
${a}_{5} = {a}_{4} + 2 \left(5\right)$
${a}_{5} = - 1 + 10$
${a}_{5} = 9$

The first five terms of the sequence are $4 , - 3. 5 , - 1$ and $9$.

Nov 9, 2016

$4 , - 3 , 5 , - 1 , 9$

with general formula: ${a}_{n} = \frac{1}{6} \left({2}^{n}\right) - \frac{11}{3} {\left(- 1\right)}^{n}$

#### Explanation:

${a}_{1} = 4$

${a}_{2} = - 3$

${a}_{3} = {a}_{2} + 2 {a}_{1} = - 3 + 2 \left(4\right) = 5$

${a}_{4} = {a}_{3} + 2 {a}_{2} = 5 + 2 \left(- 3\right) = - 1$

${a}_{5} = {a}_{4} + 2 {a}_{3} = - 1 + 2 \left(5\right) = 9$

$\textcolor{w h i t e}{}$
Bonus

How about a general non-recursive formula for ${a}_{n}$?

Suppose the series is asymptotic to a geometric series with common ratio $r$. Then $r$ must satisfy:

${r}^{2} = r + 2$

That is:

$0 = {r}^{2} - r - 2 = \left(r - 2\right) \left(r + 1\right)$

So $r = 2$ or $r = - 1$

Thus we find that the geometric series:

${b}_{n} = {2}^{n} \text{ }$ ($2 , 4 , 8 , 16 , 32 , 64 , \ldots$)

${c}_{n} = {\left(- 1\right)}^{n} \text{ }$ ($- 1 , 1 , - 1 , 1 , - 1 , 1 , \ldots$)

both conform to the formula: ${a}_{n + 2} = {a}_{n + 1} + 2 {a}_{n}$

Note that any linear combination of ${b}_{n}$ and ${c}_{n}$ will also satisfy the same rule.

So look for a general formula of the form:

${a}_{n} = {2}^{n} A + {\left(- 1\right)}^{n} B$

Putting $n = 1$ we have:

$4 = {a}_{1} = {2}^{1} A + {\left(- 1\right)}^{1} B = 2 A - B$

Putting $n = 2$ we have:

$- 3 = {a}_{2} = {2}^{2} A + {\left(- 1\right)}^{2} B = 4 A + B$

Adding these two equations, we find:

$1 = 6 A$

So $A = \frac{1}{6}$ and $B = - 3 - 4 A = - 3 - \frac{2}{3} = - \frac{11}{3}$

So:

${a}_{n} = \frac{1}{6} \left({2}^{n}\right) - \frac{11}{3} {\left(- 1\right)}^{n}$