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2 years ago

Find the first four terms of the recursive sequence defined by the following formula: an = an-1 4 where a4 = 2 1 4 , , , 2 1 4.

1 answer:

7 0

The first four-term of the sequence is found by the given equation $a_{n-1} = 4a_n$ is given as 144, 36, 9, and 9/4.

<h3>What is a sequence?</h3>

A sequence is a list of elements that have been ordered in a sequential manner, such that members come either before or after.

The given equation will be

$\rm a_n = \dfrac{a_{n-1}}{4} \\\\a_{n-1} = 4a_n$

And

$a_4 = 2 \dfrac{1}{4} = \dfrac{9}{4}$

For n = 4, we have

$\rm a_{4-1} = 4a_4\\\\a_3 \ \ \ = 4*\dfrac{9}{4} \\\\ a_3 \ \ \ = 9$

For n = 3, we have

$\rm a_{3-1} = 4a_3\\\\a_2 \ \ \ = 4*9 \\\\ a_2 \ \ \ = 36$

For n = 2, we have

$\rm a_{2-1} = 4a_2\\\\a_1 \ \ \ = 4*36 \\\\ a_1 \ \ \ = 144$

More about the sequence link is given below.

brainly.com/question/21961097

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Can someone give me the answer and how you work this? -5 - (-10) = ?

erastovalidia [21]

Answer:

Step-by-step explanation:

-5 - (-10)

Here we have to use sign of multiplication rule.

negative * negative = positive

that is, -(-) = +

We use this rule and simplify.

- (-10) = + 10

Therefore, -5 + 10

When we have different sign of the numbers, we have to subtract from greatest to least number, then put the largest number sign in the answer.

-5 + 10 = + 5

Hence, the answer is 5

Hope this is will help you.

Thank you.

7 0

2 years ago

Evaluate using integration by parts

PolarNik [594]

Rather than carrying out IBP several times, let's establish a more general result. Let

$I(n)=\displaystyle\int x^ne^x\,\mathrm dx$

One round of IBP, setting

$u=x^n\implies\mathrm du=nx^{n-1}\,\mathrm dx$

$\mathrm dv=e^x\,\mathrm dx\implies v=e^x$

gives

$\displaystyle I(n)=x^ne^x-n\int x^{n-1}e^x\,\mathrm dx$

$I(n)=x^ne^x-nI(n-1)$

This is called a power-reduction formula. We could try solving for $I(n)$ explicitly, but no need. $n=5$ is small enough to just expand $I(5)$ as much as we need to.

$I(5)=x^5e^x-5I(4)$

$I(5)=x^5e^x-5(x^4e^x-4I(3))=(x-5)x^4e^x+20I(3)$

$I(5)=(x-5)x^4e^x+20(x^3e^x-3I(2))=(x^2-5x+20)x^3e^x-60I(2)$

$I(5)=(x^2-5x+20)x^3e^x-60(x^2e^x-2I(1))=(x^3-5x^2+20x-60)x^2e^x+120I(1)$

$I(5)=(x^3-5x^2+20x-60)x^2e^x+120(xe^x-I(0))$

Finally,

$I(0)=\displaystyle\int e^x\,\mathrm dx=e^x+C$

so we end up with

$I(5)=(x^4-5x^3+20x^2-60x+120)xe^x-120e^x+C$

$I(5)=(x^5-5x^4+20x^3-60x^2+120x-120)e^x+C$

and the antiderivative is

$\displaystyle\int2x^5e^x\,\mathrm dx=(2x^5-10x^4+40x^3-120x^2+240x-240)e^x+C$

8 0

2 years ago

HELP AGAIN!!! Plzzzzz thank you!!!

never [62]

Absolute value of -1/3 is 1/3

6 0

3 years ago

after 2/5 of the kittens were given away, there was k kittens and p puppies left in the pet store. write and algebraic expressio

Rufina [12.5K]

Answer:

Algebraic expression in terms of k

k= T- p

Step-by-step explanation:

no of kittens left = k

no of puppies left =p

If total left after giving away kittens =T

T = p + k

In terms of k the expression will be

k= T- p

Keywords: Algebra

Learn more about algebra at:

#learnwithBrainly

6 0

3 years ago

According to a recent survey, the population distribution of number of years of education for self-employed individuals in a c

creativ13 [48]

Answer:

a) X: number of years of education

b) Sample mean = 13.5, Sample standard deviation = 0.4

c) Sample mean = 13.5, Sample standard deviation = 0.2

d) Decrease the sample standard deviation

Step-by-step explanation:

We are given the following in the question:

Mean, μ = 13.5 years

Standard deviation,σ = 2.8 years

a) random variable X

X: number of years of education

Central limit theorem:

If large random samples are drawn from population with mean $\mu$ and standard deviation $\sigma$ , then the distribution of sample mean will be normally distributed with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

b) mean and the standard for a random sample of size 49

$\mu_{\bar{x}} = \mu = 13.5\\\\\sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{2.8}{\sqrt{49}} = 0.4$

c) mean and the standard for a random sample of size 196

$\mu_{\bar{x}} = \mu = 13.5\\\\\sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{2.8}{\sqrt{196}} = 0.2$

d) Effect of increasing n

As the sample size increases, the standard error that is the sample standard deviation decreases. Thus, quadrupling sample size will half the standard deviation.

7 0

3 years ago