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marissa [1.9K]
3 years ago
7

How to find the first n terms of a quadratic sequence a1 = 3, r = 2, n =5

Mathematics
1 answer:
Nezavi [6.7K]3 years ago
3 0

We suspect you mean "geometric sequence". Multiply each term by r to get the next.

a1 = 3 . . . . . given

a2 = 3*2 = 6

a3 = 6*2 = 12

a4 = 12*2 = 24

a5 = 24*2 = 48

_____

A <em>quadratic</em> sequence is one in which second-differences are a constant. You have not provided enough information about the sequence so that more terms of a quadratic sequence could be found. While you may occasionally see problems asking you to write the formula for one of these, they are rarely studied in any detail in Algebra. At least three terms are needed before the next terms can be predicted.

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Answer:

a) f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

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Step-by-step explanation:

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f'(t)=0.001155t(t-1980)^{0.5}

a) To find the function f(t) you integrate f(t):

\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt

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\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}

Next, you replace in the integral:

\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C

Then, the function f(t) is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient​ visits.

Hence C' = 264,034,000

The function is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

b) For t = 2015 you have:

f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7

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