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Softa [21]
3 years ago
10

Find the first five terms of the sequence given the following recursive formula:

Mathematics
1 answer:
forsale [732]3 years ago
5 0

Answer:

The first five terms are;

-3,-7,11,-29,69

Step-by-step explanation:

The recursive definition of the sequence is

a_1=-3,, a_2=-7 and a_n=a_{n-2}-2a_{n-1}.

When n=3, we obtain;

a_3=a_{3-2}-2a_{3-1}.

\implies a_3=a_{1}-2a_{2}.

\implies a_3=-3-2(-7).

\implies a_3=11.

When n=4

\implies a_4=a_{2}-2a_{3}.

\implies a_4=-7-2(11)=-29.

When n=5

\implies a_5=a_{3}-2a_{4}.

\implies a_4=11-2(-29)=69.

Therefore the first five terms are;

-3,-7,11,-29,69

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NEED ASAP plz. I will also mark brainiest! Don't just take points, I will report you.
PilotLPTM [1.2K]

Part A:

You may choose the two lines connecting the origin and points A and B, and choose the portion of the space between them.

The line between the origin and A is

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We want everything below this line (line included), so the first inequality is

y \leq 3x

The line between the origin and B is

y = \dfrac{1}{3}x

We want everything above this line (line included), so the second inequality is

y \geq \dfrac{1}{3}x

Create a system with these two inequalities and you'll have an area including only points A and B

Part B:

To verify the solutions, we can plug the coordinates of A and B in this system and check that we get something true: the coordinates of point A are (1,3), while the coordinates of point B are (3,1). The system becomes:

A:\begin{cases}3 \leq 3\cdot 1\\3 \geq \frac{1}{3}\cdot 1\end{cases},\quad B:\begin{cases}1 \leq 3\cdot 3\\1 \geq \frac{1}{3}\cdot 3\end{cases}

Which means

A:\begin{cases}3 \leq 3\\3 \geq \frac{1}{3}\end{cases},\quad B:\begin{cases}1 \leq 9\\1 \geq 1\end{cases}

And these are all true. So, the system is satisfied, which means that the points belong to the shaded area.

Part C

If you draw the line, you'll see that the only points that lay below the line are B and C. In fact, if we plug the coordinates we have

B:\ 1

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