## 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?

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.

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.

## 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!