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Gnesinka [82]
3 years ago
8

Given the infinite series:

5%29%7D%2B%5Cfrac%7B1%7D%7B%285%29%287%29%7D%2B..." id="TexFormula1" title="\frac{1}{(1)(3)} +\frac{1}{(3)(5)}+\frac{1}{(5)(7)}+..." alt="\frac{1}{(1)(3)} +\frac{1}{(3)(5)}+\frac{1}{(5)(7)}+..." align="absmiddle" class="latex-formula">
a. Rewrite the series in sigma notation
b. Write an explicit formula for the nth partial sum
Mathematics
1 answer:
chubhunter [2.5K]3 years ago
8 0

a) The infinite series in <em>sigma</em> notation is described by this expression:

y = \sum \limits_{i=1}^{\infty} \frac{1}{i\cdot (i+2)}     (1)

b) The <em>explicit</em> formula for the n-th <em>partial</em> sum is represented by the following expression:

y = \frac{1}{i\cdot (i+2)}, i ∈ \mathbb{N}     (2)

<h3>How to derive an expression for a monotonous series</h3>

An infinite series is <em>monotonous</em> when it is <em>bounded</em>, that is, when the limit of the <em>infinite</em> series exists. In this case, we have an evidence of monotony in the denominators of the terms of the given series. In two consecutive terms, the latter always have a denominator greater than the former.

a) The series in <em>sigma</em> notation is now described below:

y = \sum \limits_{i=1}^{\infty} \frac{1}{i\cdot (i+2)}

b) The <em>explicit</em> formula for the n-th <em>partial</em> sum is defined by the expression within the sum, which is now presented below:

y = \frac{1}{i\cdot (i+2)}, i ∈ \mathbb{N}  

To learn more on infinite series: brainly.com/question/4268280

#SPJ1

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Wally's T-shirt company estimates the demand function for the Red sox world series t-shirt to be:
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Answer:costC(p)= 1400+20p

R(p)= 4p^2+200p

Profit= 4p^2+ 180p-1400

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q=4p+200

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The sum of the first number cubed and the second number is 500 and the product is a maximum.
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We are given the following information:

the sum of the first number cubed and the second number is 500

their product is a maximum

We are looking for the 2 missing numbers.

To answer this, let's represent the two numbers as x and w.

From the given, we can form the following equation:

x^3+w=500

We can then express y as:

w=500-x^3

We can express their product as:

x(500-x^3)=500x-x^4

To find the maximum value of x, let's solve for the derivative of -x^4 + 500 x.

\begin{gathered} f(x)=-x^4+500x \\ f^{\prime}(x)=-4x^3+500 \end{gathered}

Then we solve for the value of x where f'(x) = 0.

\begin{gathered} -4x^3+500=0 \\ -4x^3=-500 \\ x^3=125 \\ x=5 \end{gathered}

Then we use x = 5 to solve for the second number, w.

\begin{gathered} w=500-x^3 \\ w=500-5^3 \\ w=500-125 \\ w=375 \end{gathered}

Therefore, the two numbers are 5 and 375.

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- 1 +  \sqrt{3}

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