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MAXImum [283]
3 years ago
8

You write 4 different names on index cards.You shuffle the cards and then lay them out in a row. What is probability that they w

ill come out in alphabetical order? Express your answer as a fraction in simplest form.
1/256


1/24


1/4


1/6
Mathematics
1 answer:
MA_775_DIABLO [31]3 years ago
3 0
It would be B- 1/24. 

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Fynjy0 [20]

The sum we want is

\displaystyle \sum_{n=0}^\infty \frac{(-1)^{T_n}}{(2n+1)^2} = 1 - \frac1{3^2} - \frac1{5^2} + \frac1{7^2} + \cdots

where T_n=\frac{n(n+1)}2 is the n-th triangular number, with a repeating sign pattern (+, -, -, +). We can rewrite this sum as

\displaystyle \sum_{k=0}^\infty \left(\frac1{(8k+1)^2} - \frac1{(8k+3)^2} - \frac1{(8k+7)^2} + \frac1{(8k+7)^2}\right)

For convenience, I'll use the abbreviations

S_m = \displaystyle \sum_{k=0}^\infty \frac1{(8k+m)^2}

{S_m}' = \displaystyle \sum_{k=0}^\infty \frac{(-1)^k}{(8k+m)^2}

for m ∈ {1, 2, 3, …, 7}, as well as the well-known series

\displaystyle \sum_{k=1}^\infty \frac{(-1)^k}{k^2} = -\frac{\pi^2}{12}

We want to find S_1-S_3-S_5+S_7.

Consider the periodic function f(x) = \left(x-\frac12\right)^2 on the interval [0, 1], which has the Fourier expansion

f(x) = \frac1{12} + \frac1{\pi^2} \sum_{n=1}^\infty \frac{\cos(2\pi nx)}{n^2}

That is, since f(x) is even,

f(x) = a_0 + \displaystyle \sum_{n=1}^\infty a_n \cos(2\pi nx)

where

a_0 = \displaystyle \int_0^1 f(x) \, dx = \frac1{12}

a_n = \displaystyle 2 \int_0^1 f(x) \cos(2\pi nx) \, dx = \frac1{n^2\pi^2}

(See attached for a plot of f(x) along with its Fourier expansion up to order n = 10.)

Expand the Fourier series to get sums resembling the S'-s :

\displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{k=0}^\infty \frac{\cos(2\pi(8k+1) x)}{(8k+1)^2} + \sum_{k=0}^\infty \frac{\cos(2\pi(8k+2) x)}{(8k+2)^2} + \cdots \right. \\ \,\,\,\, \left. + \sum_{k=0}^\infty \frac{\cos(2\pi(8k+7) x)}{(8k+7)^2} + \sum_{k=1}^\infty \frac{\cos(2\pi(8k) x)}{(8k)^2}\right)

which reduces to the identity

\pi^2\left(\left(x-\dfrac12\right)^2-\dfrac{21}{256}\right) = \\\\ \cos(2\pi x) {S_1}' + \cos(4\pi x) {S_2}' + \cos(6\pi x) {S_3}' + \cos(8\pi x) {S_4}'  \\\\ \,\,\,\, + \cos(10\pi x) {S_5}' + \cos(12\pi x) {S_6}' + \cos(14\pi x) {S_7}'

Evaluating both sides at x for x ∈ {1/8, 3/8, 5/8, 7/8} and solving the system of equations yields the dependent solution

\begin{cases}{S_4}' = \dfrac{\pi^2}{256} \\\\ {S_1}' - {S_3}' - {S_5}' + {S_7}' = \dfrac{\pi^2}{8\sqrt 2}\end{cases}

It turns out that

{S_1}' - {S_3}' - {S_5}' + {S_7}' = S_1 - S_3 - S_5 + S_7

so we're done, and the sum's value is \boxed{\dfrac{\pi^2}{8\sqrt2}}.

6 0
2 years ago
Write the equation in standard form. Identify the important features of the graph:
Lerok [7]

Answer:

Standard form: \left(x-\dfrac{9}{2}\right)^2+(y+5)^2=\dfrac{121}{4}.

This equation represents the circle with the center at the point \left(\dfrac{9}{2},-5\right) and the radius r=\dfrac{11}{2}.

Step-by-step explanation:

Consider expression x^2+y^2-9x+10y+15=0.

First, form perfect squares:

(x^2-9x)+(y^2+10y)+15=0,\\ \\\left(x^2-9x+\dfrac{81}{4}\right)-\dfrac{81}{4}+(y^2+10y+25)-25+15=0,\\ \\\left(x-\dfrac{9}{2}\right)^2+(y+5)^2=10+\dfrac{81}{4},\\ \\\left(x-\dfrac{9}{2}\right)^2+(y+5)^2=\dfrac{121}{4}.

This equation represents the circle with the center at the point \left(\dfrac{9}{2},-5\right) and the radius r=\dfrac{11}{2}.

6 0
3 years ago
Joshua uses a rule to write the following sequence of numbers 1/6,1/2,5/6,----------,11/2 What rule did Joshua use? What is the
Pani-rosa [81]

Answer: The missing number in the sequence is \frac{7}{6}

Step-by-step explanation:

Since we have given that

\frac{1}{6},\frac{1}{2},\frac{5}{6},----------,\frac{11}{2}

First term = a= \frac{1}{6}

Common difference = d is given by

d=a_2-a_1\\\\\frac{1}{2}-\frac{1}{6}\\\\=\frac{6-2}{12}\\\\=\frac{4}{12}=\frac{1}{3}\\\\Similarly,\\d=a_3-a_2\\\\=\frac{5}{6}-\frac{1}{2}\\\\=\frac{10-6}{12}\\\\=\frac{4}{12}\\\\=\frac{1}{3}

Therefore, it forms an arithmetic sequence.

Since, a_4 is missing,

So,

a_4=a+3d\\\\a_4=\frac{1}{6}+3\times \frac{1}{3}\\\\a_4=\frac{1}{6}+1\\\\a_4=\frac{1+6}{6}\\\\a_4=\frac{7}{6}

Hence, the missing number in the sequence is \frac{7}{6}

4 0
3 years ago
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3 years ago
Please, with work, find the prime factorization of 60 and 140?
Brrunno [24]
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3 0
3 years ago
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