Find the 58th term of the following arithmetic sequence. 13, 19, 25, 31,

Mathematics

1 answer:

pashok25 [27]2 years ago

6 0

Answer:

the answer to this is 355.

source:da brain

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Find the number to which the sequence {(3n+1)/(2n-1)} converges and prove that your answer is correct using the epsilon-N defini

Nat2105 [25]

By inspection, it's clear that the sequence must converge to $\dfrac32$ because

$\dfrac{3n+1}{2n-1}=\dfrac{3+\frac1n}{2-\frac1n}\approx\dfrac32$

when $n$ is arbitrarily large.

Now, for the limit as $n\to\infty$ to be equal to $\dfrac32$ is to say that for any $\varepsilon>0$ , there exists some $N$ such that whenever $n>N$ , it follows that

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|$

From this inequality, we get

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|=\left|\dfrac{(6n+2)-(6n-3)}{2(2n-1)}\right|=\dfrac52\dfrac1{|2n-1|}$
$\implies|2n-1|>\dfrac5{2\varepsilon}$
$\implies2n-1\dfrac5{2\varepsilon}$
$\implies n\dfrac12+\dfrac5{4\varepsilon}$

As we're considering $n\to\infty$ , we can omit the first inequality.

We can then see that choosing $N=\left\lceil\dfrac12+\dfrac5{4\varepsilon}\right\rceil$ will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that $N\in\mathbb N$ .

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sammy [17]

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elixir [45]

The Answer is C 5/6

250/300 reduced and simplified is 5/6

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Type the correct answer in the box. Use numerals instead of words. Consider this expression. Vas – 7+ 161 When a = 2 and b = -4,

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Correct Question:

Type the correct answer in the box. Use numerals instead of words. Consider this expression. $\sqrt{a^2} - 7 + |b|$ When a = 2 and b = -4, the value of the expression is