precalculus algebra – Factoring in an induction test

I have to try the following:

$ 1 + 3 ^ 3 + … + (2n + 1) ^ 3 = (n + 1) ^ 2 (2n ^ 2 + 4n + 1) $ Inductively.

My intent:

Base case, $ n = 1 $:

$ 1 + 3 ^ 3 = (2) ^ 2 (2 cdot1 ^ 2 + 4 cdot1 + 1) $, which is true.

By inductive hypothesis, suppose $ n = k $:

$ 1 + 3 ^ 3 + … + (2k + 1) ^ 3 = (k + 1) ^ 2 (2k ^ 2 + 4k + 1) $

by $ n = k + 1 $

$ 1 + 3 ^ 3 + … + (2k + 1) ^ 3 + (2k + 3) ^ 3 = (k + 2) ^ 2 (2 (k + 1) ^ 2 + 4 (k + 1) + 1) $

Using the inductive hypothesis we need to prove that

$ (k + 1) ^ 2 (2k ^ 2 + 4k + 1) + (2k + 3) ^ 3 = (k + 2) ^ 2 (2 (k + 1) ^ 2 + 4 (k + 1) + 1) $

And here I have the problem, because I do not know how to manipulate any of the sides of the equation to prove this. I tried it on both sides but I can not find the way. One of my last attempts ended here:

$ (k + 1) ^ 2 (2k ^ 2 + 4k + 1) + (2k + 3) ^ 3 = (k + 2) ^ 2 (2 (k + 1) (k + 3) + 1) $

Which is true for alpha wolfram.

PS: I am aware that there are many documents and information about this test, but I am not looking for another form or something, because I saw many publications on how to prove this claim through induction, but all were used. the last term is (2n-1) ^ 3, and I need to prove it when it is (2n + 1) ^ 3, and the final expression is a bit different. I just need help factoring my last step.

Corrections in the inductive steps that I followed are also appreciated.

Precalculus of algebra – Polynomial expansion (trick plus / minus in statistics)

Let's suppose the random variables $ X_1, …, X_n $ They are independent and identically distributed. Leave $ mu_x $ denotes the expected value of $ X $. I often see the following plus / minus trick used in statistics, that is,

begin {align *} frac {1} {n} sum_ {i = 1} ^ n (X_i – bar {X}) ^ 2 & = frac {1} {n} sum_ {i = 1 } ^ n (X_i – mu_X + mu_X – bar {X}) ^ 2 \
& = frac {1} {n} sum_ {i = 1} ^ n (X_i – mu_x) ^ 2 – ( bar {X} – mu_x) ^ 2 end {align *}

Can someone formulate this in an identity for me? I mean, something like

$ (a – b) ^ 2 = (a – c + c – b) ^ 2 = (a – c) ^ 2 + (c – b) ^ 2 + text {some crossed term}? $

What exactly is that term crossed?

In a similar token, I've also seen
begin {align *} frac {1} {n} sum_ {i = 1} ^ n (X_i – bar {X}) (Y_i – bar {Y}) & = frac {1} {n } sum_ {i = 1} ^ n (X_i – mu_X) (Y_i – mu_Y) – ( bar {X} – mu_x) ( bar {Y} – mu_Y) end {align *}

What identity of polynomial expansion is at stake here?

Precalculus of algebra – Is there any way to maximize the given function?

While working on a question, I came across the following equation

$$ c = frac {(2-a_0) + sqrt {-a_0 (3a_0-4)}} {2} $$

We are obliged to maximize the function. $ c $ and find its corresponding value

I drew his graph in Desmos and found that his maximum was 1/3, however I have no idea how to get to this result.

Nor are we allowed to use calculus (at least it discourages you)
How are we supposed to proceed with this?

Precalculus of algebra: why is not there a solution set for $ | x-7 |

They ask me to find the set of solutions for $ | x-7 | <-4 $?

I arrived at $ (- infty, 3) cup (11, infty) $

by $ x – 7> 0 $:

$ x-7 <-4 $ = $ x <3 $

by $ x-7 <0 $ = $ – (x-7) <- 4 $ = $ -x + 7 <-4 $ = $ -x <-11 $ = $ x> $ 11

Then, I come to a solution of:

$ (- infty, 3) cup (11, infty) $

However, my textbook says "there is no solution". Why is there no solution?

precalculus algebra: solve the equation on reals: $ sqrt {5x ^ 2 + 27x + 25} – 5 sqrt {x + 1} = sqrt {x ^ 2 – 4} $.

Solve the equation over reais: $ sqrt {5x ^ 2 + 27x + 25} – 5 sqrt {x + 1} = sqrt {x ^ 2 – 4} $.

This problem is an adaptation of a recent competition. And I can not solve it.

The solutions are $ sqrt {5} + 1 $ Y $ dfrac {13 + sqrt {65}} {8} $, as Wolfram Alpha says.

precalculus algebra – Solve fr x y $ x ^ 2 + y ^ 2 = 1 $ y $ x pm y = frac pi4 $

Solve fr x y $ x ^ 2 + y ^ 2 = 1 $ Y $ x pm y = frac pi4 $

I tried to solve this by a substitute method. And using the quadratic formula, but that creates many cases.
The original problem was to solve. $ arcsin x + arcsin y = frac pi2 $ Y $ without 2x = cos2y $.

Precalculus of algebra: how to calculate the area of ​​the triangle with perpendicular lines

  1. In the following figure, the lines have slopes of 3 and 5. The lines intersect at (10,15). How far is it between the x-intercepts of the lines?

enter the description of the image here

The equation of the line with slope of 3 is $ y = 3x + b $ and with the point $ (10,15) $ is $ 15 = 3 (10) + b $ then the equation is $ y = 3x-15 $

Similarly, the slope is 5. Then the equation is $ y = 5x-35 $

The x interceptions are $ x = 5 $ for $ y = 3x-15 $ Y $ x = 7 $ for $ y = 5x-35 $. The distance between the two 2. Is this correct?

  1. In the following figure, the two lines are perpendicular, and intersect in $ (6.8) $. The intersections in and of the lines have a sum of zero. Find the area of ​​the shaded region.

enter the description of the image here

the equations of the lines are $ y = mx + b $ Y $ y = – ( frac {1} {m}) x-b $ since the intersections in and are opposite each other. If I reconnect the point, I have:

$ 8 = 6m + b $ Y $ 8 = – ( frac {1} {m}) (6) -b $

If I put these two equations together, I would have

$ 6m + b = – ( frac {1} {m}) (6) -b $

$ 6m = – frac {6} {m} $

$ 6m ^ 2 = -6 $

$ 6m ^ 2 + 6 = 0 $

$ 6 (m ^ 2 + 1) = 0 $

Here I am stuck because I have a negative root, so I'm pretty sure I tried this problem badly.

Precalculus of algebra: interrelated constraints through linear combinations

Dice $ x $ Y $ and $ Real variables are such that:

$ left | x right | le alpha, left | and right | le alpha, $

where $ alpha $ It is a positive constant.
I want to determine the limits of $ u, v $ where $ u, v $ are determined by a combination of $ x, y $ as follows:
begin {align}
& u text {} = text {} x text {} – text {} y, \
& v text {} = text {} y. \
end {align}

My approach is:
begin {align}
& -2 alpha le u = x text {} – text {} and le 2 alpha, \
& – alpha le v text {} = text {} and le alpha. \
end {align}

Therefore, the limits of $ u, v $ are :
$ left | u right | le 2 , left | v right | le alpha, $

I tried to verify my result by transforming the original equation into:
begin {align}
& x text {} = text {} u text {} + text {} v, \
& y text {} = text {} v. \
end {align}

Then using the derived restrictions, we can conclude
$ left | x right | le 3 , left | and right | le alpha, $

which is different from the original statement.
Could you please help me indicate where my error is? And how to solve this problem?

Precalculus algebra – Order to prove inequalities

at some point when I'm trying inequalities
Example
For positive a, b, c, d and $ abcd = 16 $
Show that $ ( frac {a} {b}) ^ 3 + ( frac {b} {c}) ^ {3} + ( frac {c} {d}) ^ 3 + ( frac {d} {a }) ^ 3 + 4 geq a + b + c + d $

My test But I can not finish …

First, L.H.S we use the power, the inequalities that we will obtain.

$ LHS geq ( frac { frac {a} {b} + frac {b} {c} + frac {c} {d} + frac {d} {a}} {16}) ^ { 3} +4 $
And R.H.S. I multiply with $ frac {2} { sqrt[4]{abcd}} $ Now the inequality is homogeneous, I will ignore the condition abcd = 16

We have to show that $ ( frac { frac {a} {b} + frac {b} {c} + frac {c} {d} + frac {d} {a}} {16}) ^ {3} + 4 geq frac {2 (a + b + c + d)} { sqrt[4]{abcd}} $

From A.M.-G.M; $ sum_ {cyc}
frac {a} {b} geq 4 $
…. so at the end

I have $ 4 sqrt[4]{abcd} geq a + b + c + d $ But following with wrong inverse sign ….

So, is there a trick about what I should do first to prove the inequality? A.M-G.M or Cauchy ….. that is
Thank you !

Precalculus of algebra – Is there a mathematical function for x + x ^ 2 + x ^ 3 + x ^ 4 + x ^ 5 …. etc?

I am a UI programmer and I am trying to calculate an effect related to mouse movement. The behavior that I'm noticing is equivalent to x + x ^ 2 + x ^ 3 + x ^ 4 + x ^ 5 ... and so on.

For the purposes of my application, I just have to go to Like x ^ 3 before the result is close enough, but I wonder if there is a mathematical function or operator, ideally a version in common programming languages, designed for this type Of what has more precision.

Thank you!