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2

## F is a transfomation of R^3 to R^3 so that f(x,y,z)=(x-y,x+y,z-x) How can i find the standart matrix of this transformation and how can i find the inverse of this transformation? I am not sure how to approach this problem

Geoff K.
Featured 1 month ago

The matrix is F=[(1,–1,0),(1,1,0),(–1,0,1)].

The inverse matrix is F^(–1)=[(1/2,1/2,0),(–1/2,1/2,0),(1/2,1/2,1)].

#### Explanation:

A linear transformation such as

$f \left(x , y , z\right) = \left(x - y , \text{ "x+y," } z - x\right)$

can be expressed as matrix multiplication. Just like how the function $f$ takes in the vector $\stackrel{\rightarrow}{x} = \left(x , y , z\right)$ and gives back a new vector $\left(x - y , \text{ "x+y," } z - x\right) ,$ we are looking for the $3 \times 3$ matrix $F$ that satisfies

$F \times \stackrel{\rightarrow}{x} = \left[\begin{matrix}x - y \\ x + y \\ z - x\end{matrix}\right]$.

Let's remember how matrix multiplication works: when multiplying a $3 \times 3$ matrix $\left[\begin{matrix}a & b & c \\ d & e & f \\ g & h & i\end{matrix}\right]$ with a $3 \times 1$ vector $\left[\begin{matrix}x \\ y \\ z\end{matrix}\right] ,$ the result is the $3 \times 1$ vector as follows:

${\left[\begin{matrix}a & b & c \\ d & e & f \\ g & h & i\end{matrix}\right]}_{\textcolor{red}{3 \times 3}} \times {\left[\begin{matrix}x \\ y \\ z\end{matrix}\right]}_{\textcolor{red}{3 \times 1}} = {\left[\begin{matrix}a x + b y + c z \\ \mathrm{dx} + e y + f z \\ g x + h y + i z\end{matrix}\right]}_{\textcolor{red}{3 \times 1}}$

We want to find what $a , b , c , \ldots , i$ need to be so that the product vector (on the right) matches the output vector from the function $f .$

In other words, we want $\left[\begin{matrix}a x + b y + c z \\ \mathrm{dx} + e y + f z \\ g x + h y + i z\end{matrix}\right] = \left[\begin{matrix}x - y \\ x + y \\ z - x\end{matrix}\right] .$

It may still look like a jumble of letters, but remember—we're going to choose what $a$ through $i$ are. Let's look at the first line of the equality:

$a x + b y + c z = x - y$

What do $a , b ,$ and $c$ have to be to make this true? Well, on the right we have $1 x$ and –1y, so to make the left side match, we should choose $a = 1 ,$ b=–1, and $c = 0 :$

1x+(–1)y+0z=x-y

The process is the same for the other two rows. The matrix is

F=[(1,–1,0),(1,1,0),(–1,0,1)].

Notice:

the first row contains the coefficients of $x - y ,$
the second row contains the coefficients of $x + y ,$ and
the third row contains the coefficients of $z - x ,$

all in $x , y , z$ order, of course.

The inverse matrix F^(–1) is the matrix which "undoes" the action done by $F .$ That is, just as $F$ takes the input $\stackrel{\rightarrow}{x}$ and gives $F \stackrel{\rightarrow}{x}$, the matrix F^(–1) takes in $F \stackrel{\rightarrow}{x}$ as input and gives back just $\stackrel{\rightarrow}{x} :$

F^(–1) xx Fstackrelrarrx = stackrelrarrx

or

F^(–1) xx [(x-y),(x+y),(z-x)] = [(x),(y),(z)].

You can find this inverse the same way as before; this one's not too hard to visualize, though others may take some effort. You can also find it by computing the inverse of $F .$ Instructions should be available in any introductory matrix algebra textbook.

F^(–1)=[(1/2,1/2,0),(–1/2,1/2,0),(1/2,1/2,1)]

2

## How to sketch the graph f(x)= |x^2-1|/(x^2-1) showing any points of discontinuity, horizontal or vertical asymptotes and x and y intercepts?

Jim H
Featured 2 months ago

First note that $\frac{\left\mid u \right\mid}{u} = \left\{\begin{matrix}1 & \text{if" & u > 0 \\ -1 & "if} & u < 0\end{matrix}\right.$

#### Explanation:

Second observe that ${x}^{2} - 1 > 0$ if and only if $x < - 1$ OR $x > 1$.

So we have

$f \left(x\right) = \frac{\left\mid {x}^{2} - 1 \right\mid}{{x}^{2} - 1} = \left\{\begin{matrix}1 & \text{if" & x < -1 \\ -1 & "if" & -1 < x < 1 \\ 1 & "if} & 1 < x\end{matrix}\right.$

So the graph is:

graph{abs(x^2-1)/(x^2-1) [-4.995, 4.87, -1.794, 3.14]}

It is now clear that there are
two discontinuities (jump),
no vertical asymptotes,
no $x$-intercepts,
$y$-intercept is $- 1$ and,
depending on your definition of horizontal asymptote, either none or $y = 1$ on both the left and right.

2

## If the point (0,0) is on the graph of y=f(x), which of the following points must be on the graph of y=f(x-2)+3? a)(2, -3) b)(2,2) c)(2,3) d)(-2,-3) e)(-2,3)

Jake M.
Featured 2 months ago

C

#### Explanation:

So let's do this two ways: using transformations of a function, and using the logic behind those transformations.

Transformations
We know that the transformation $x \rightarrow x - 2$ shifts all points 2 to the right while transforming $y \rightarrow y + 3$ shifts everything up three, hence (0,0) becomes (2, 3) i.e. c.

Logic
The only thing we know about the function $f \left(x\right)$ is that $f \left(0\right) = 0$.
Well, if we want to mess with the given equation for y
$y = f \left(x - 2\right) + 3$
so that the argument of $f$ is 0, we obviously need $x - 2 = 0$ or $x = 2$.
When we plug in the value for $f \left(0\right)$ i.e. 0, we see that $y = 0 + 3 = 3$, as shown before.

3

## What is the relationship between the roots and coefficients of a polynomial?

George C.
Featured 1 month ago

Manifold...

#### Explanation:

There are many things we can say about the relationship between the roots of a polynomial and its coefficients. Here are just a few...

Rational roots theorem

Given a polynomial:

${a}_{n} {x}^{n} + {a}_{n - 1} {x}_{n} - 1 + \ldots + {a}_{1} x + {a}_{0} \text{ }$

with integer coefficients and ${a}_{n} \ne 0$

any rational zeros are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor (positive or negative) of the constant term ${a}_{0}$ and $q$ a divisor of the coefficient ${a}_{n}$ of the leading term.

The same is true if we specify Gaussian integers instead of integers, or indeed the elements of any integral domain.

Descartes' Rule of Signs

For any polynomial with real coefficients written in standard form, the number of changes in the signs of the coefficients gives the maximum number of positive real zeros. If the number of zeros is less, then it is less by a multiple of $2$

For example:

${x}^{4} + {x}^{3} - 4 {x}^{2} + 2 x - 5$

has coefficients with signs $+ + - + -$. With $3$ changes, we can deduce that this quartic has $3$ or $1$ positive real zeros.

Furthermore, if you invert the signs on the terms of odd degree, then the number of changes in the resulting pattern indicates the number of negative real zeros. In our example, we would get $+ - - - -$, so with one change of sign we could tell that the quartic has exactly one negative real zero.

Discriminant

You are almost certainly familiar with the formula for the discriminant $\Delta$ of a quadratic $a {x}^{2} + b x + c$, namely:

$\Delta = {b}^{2} - 4 a c$

Assuming that $a , b , c$ are rational, we can interpret its value as follows:

• If $\Delta > 0$ is a perfect square, then the quadratic has two distinct rational zeros.

• If $\Delta > 0$ is not a perfect square, then the quadratic has two distinct irrational zeros.

• If $\Delta = 0$ then the quadratic has one repeated rational zero.

• If $\Delta < 0$ then the quadratic has two distinct non-real complex zeros which are complex conjugates of one another.

Polynomials of higher degree also have discriminants. They are not particularly useful for quartics and above, except to identify when they have repeated zeros, but for cubics they are very useful.

The discriminant of $a {x}^{3} + b {x}^{2} + c x + d$ is given by:

$\Delta = {b}^{2} {c}^{2} - 4 a {c}^{3} - 4 {b}^{3} d - 27 {a}^{2} {d}^{2} + 18 a b c d$

• If $\Delta > 0$ the cubic has $3$ real zeros.

• If $\Delta = 0$ the cubic has a repeated zero and all three zeros are real.

• If $\Delta < 0$ the cubic has one real zero and two complex conjugate non-real zeros.

3

## What is the relationship between the roots and coefficients of a polynomial?

George C.
Featured 1 month ago

A few more...

#### Explanation:

Elementary symmetric polynomials

The coefficients of a monic polynomial are (modulo alternating signs), the elementary symmetric polynomials in the zeros.

For example:

$\left(x - \alpha\right) \left(x - \beta\right) = {x}^{2} - \left(\alpha + \beta\right) x + \alpha \beta$

$\left(x - \alpha\right) \left(x - \beta\right) \left(x - \gamma\right) = {x}^{3} - \left(\alpha + \beta + \gamma\right) {x}^{2} + \left(\alpha \beta + \beta \gamma + \gamma \alpha\right) x - \alpha \beta \gamma$

etc.

In particular, given a polynomial:

${a}_{n} {x}_{n} + {a}_{n - 1} {x}_{n - 1} + \ldots + {a}_{1} x + {a}_{0}$

the sum of its zeros is $- {a}_{n - 1} / {a}_{n}$ and the product of its zeros is ${\left(- 1\right)}^{n} {a}_{0} / {a}_{n}$.

Since any symmetric polynomial can be constructed from the elementary symmetric polynomials, we can find a polynomial that has zeros that are (say) the squares of the zeros of another polynomial - without finding what the zeros actually are.

For a substantial and important application of this, see https://socratic.org/s/aPGxwybx

Coefficient sum shortcuts

If the sum of the coefficients of a polynomial is $0$, then you can infer that $1$ is a zero.

If inverting the signs on terms of odd degree results in coefficients that sum to $0$, then you can infer that $- 1$ is a zero.

Reversing the order and reciprocals

Given a polynomial:

${a}_{n} {x}_{n} + {a}_{n - 1} {x}^{n - 1} + \ldots + {a}_{1} x + {a}_{0}$

with ${a}_{n} \ne 0$ and ${a}_{0} \ne 0$

Then the polynomial:

${a}_{0} {x}_{n} + {a}_{1} {x}_{n - 1} + \ldots + {a}_{n - 1} x + {a}_{n}$

has zeros which are reciprocals of the original polynomial.

To see why that is so, note that:

$\frac{1}{x} _ n \left({a}_{n} {x}_{n} + {a}_{n - 1} {x}^{n - 1} + \ldots + {a}_{1} x + {a}_{0}\right)$

$= {a}_{n} + {a}_{n - 1} \frac{1}{x} + \ldots + {a}_{1} \frac{1}{x} ^ \left(n - 1\right) + {a}_{0} \frac{1}{x} ^ n$

Symmetry and reciprocals

From the preceding property, we can deduce that:

If the coefficients of a polynomial are symmetrical then you can infer that the reciprocal of any zero is also a zero.

For example:

$6 {x}^{4} + 5 {x}^{3} - 38 {x}^{2} + 5 x + 6$

has zeros $2 , \frac{1}{2} , - 3 , - \frac{1}{3}$

3

## A corridor of width a meets a corridor of width b at right angles. Workmen wish to push a heavy beam on dollies around the corner, but they want to be sure it will be able to make the turn before starting. How long a beam will go around the corner ?

George C.
Featured 1 month ago

${\left({a}^{\frac{2}{3}} + {b}^{\frac{2}{3}}\right)}^{\frac{3}{2}}$

#### Explanation:

Let us locate the outer corner of the corridor at $\left(0 , 0\right)$ and the inner corner at $\left(a , b\right)$.

Consider lines of negative slope passing through $\left(a , b\right)$ and the distance between their $x$ and $y$ intercepts.

The minimum distance then models the maximum feasible length of beam.

graph{((x-14)^100+(y-13)^100-10^100)((x-20)^100+(y-20)^100-20^100)(y+x/2-5)((x-4)^2+(y-3)^2-0.02)(x^2+(y-5)^2-0.02)((x-10)^2+y^2-0.02) = 0 [-5.625, 14.375, -3, 7]}

The equation of such a line can be written in point slope form as:

$y - b = m \left(x - a\right)$

The $x$ intercept is then given by putting $y = 0$ to find:

$x = a - \frac{b}{m}$

and the $y$ intercept by putting $x = 0$ to find:

$y = b - a m$

The square of the distance between these intercepts is:

${\left(a - \frac{b}{m}\right)}^{2} + {\left(b - a m\right)}^{2} = {a}^{2} - 2 a b \frac{1}{m} + {b}^{2} \frac{1}{m} ^ 2 + {b}^{2} - 2 a b m + {a}^{2} {m}^{2}$

The minimum will occur when the derivative of this with respect to $m$ is zero. That is:

$0 = 2 a b \frac{1}{m} ^ 2 - 2 {b}^{2} \frac{1}{m} ^ 3 - 2 a b + 2 {a}^{2} m$

Multiplying by ${m}^{3} / 2$ this becomes:

$0 = a b m - {b}^{2} - a b {m}^{3} + {a}^{2} {m}^{4}$

$\textcolor{w h i t e}{0} = {a}^{2} {m}^{4} - a b {m}^{3} + a b m - {b}^{2}$

$\textcolor{w h i t e}{0} = a {m}^{3} \left(a m - b\right) + b \left(a m - b\right)$

$\textcolor{w h i t e}{0} = \left(a {m}^{3} + b\right) \left(a m - b\right)$

$\textcolor{w h i t e}{0} = \left({a}^{\frac{1}{3}} m + {b}^{\frac{1}{3}}\right) \left({a}^{\frac{2}{3}} {m}^{2} - {a}^{\frac{1}{3}} {b}^{\frac{1}{3}} m + {b}^{\frac{2}{3}}\right) \left(a m - b\right)$

Note here that the last linear factor gives $m = \frac{b}{a} > 0$, so is not suitable.

Also the quadratic factor has only non-real solutions.

So we require $m = - {b}^{\frac{1}{3}} {a}^{- \frac{1}{3}}$

With this value of $m$, the square of the distance between the intercepts is:

${\left(a - \frac{b}{m}\right)}^{2} + {\left(b - a m\right)}^{2} = {\left(a + {a}^{\frac{1}{3}} {b}^{\frac{2}{3}}\right)}^{2} + {\left(b + {a}^{\frac{2}{3}} {b}^{\frac{1}{3}}\right)}^{2}$

$\textcolor{w h i t e}{{\left(a - \frac{b}{m}\right)}^{2} + {\left(b - a m\right)}^{2}} = {a}^{\frac{2}{3}} {\left({a}^{\frac{2}{3}} + {b}^{\frac{2}{3}}\right)}^{2} + {b}^{\frac{2}{3}} {\left({b}^{\frac{2}{3}} + {a}^{\frac{2}{3}}\right)}^{2}$

$\textcolor{w h i t e}{{\left(a - \frac{b}{m}\right)}^{2} + {\left(b - a m\right)}^{2}} = {\left({a}^{\frac{2}{3}} + {b}^{\frac{2}{3}}\right)}^{3}$

So the distance is:

$\sqrt{{\left({a}^{\frac{2}{3}} + {b}^{\frac{2}{3}}\right)}^{3}} = {\left({a}^{\frac{2}{3}} + {b}^{\frac{2}{3}}\right)}^{\frac{3}{2}}$

1

## What is the polar form of ( -6,36 )?

Sridhar V.
Featured 3 hours ago

$\text{ }$
Cartesian Coordinates to Polar Form: color(blue)((-6, 36) = (36, 99.4^@)

#### Explanation:

$\text{ }$
Given the Cartesian Form: $\left(- 6 , 36\right)$

Find the Polar Form:color(blue)((r,theta)

color(green)("Step 1:"

Let us examine some of the relevant formula in context:

color(green)("Step 2:"

Plot the coordinate point color(blue)((-6, 36) on a Cartesian coordinate plane:

Indicate the known values, as appropriate:

$\overline{O A} = 6 \text{ Units}$

$\overline{A B} = 36 \text{ Units}$

Let $\overline{O B} = r \text{ Units}$

$\angle O A B = {90}^{\circ}$

Let $\angle A O B = {\alpha}^{\circ}$

color(green)("Step 3:"

Use the formula: color(red)(x^2 + y^2=r^2 to find color(blue)(r

Consider the following triangle with known values:

${r}^{2} = {6}^{2} + {36}^{2}$

$\Rightarrow 36 + 1296$

$\Rightarrow 1332$

${r}^{2} = 1332$

Hence, color(brown)(r=sqrt(1332)~~36.4966

To find the value of $\textcolor{red}{\theta}$:

$\tan \left(\theta\right) = \frac{36}{6} = 6$

$\theta = {\tan}^{-} 1 \left(6\right)$

$\theta \approx {80.53767779}^{\circ}$

color(blue)("Important:"

Since the angle color(red)(theta lines in Quadrant-II, we must subtract this angle from ${\textcolor{red}{180}}^{\circ}$ to get the required angle color(blue)(beta.

color(green)("Step 4:"

color(blue)(beta ~~ 180^@ - 80.53767779^@

$\Rightarrow \beta \approx {99.46232221}^{\circ}$

Hence, the required Polar Form:

color(blue)((r, theta) = (36, 99.4^@)

Hope it helps.

1

## 4x^3-6x is symmetric with respect to where?

Hammer
Featured 2 hours ago

The graphic of the function $x \mapsto 4 {x}^{3} - 6 x$ is symmetric with respect to the origin of the axes system.

#### Explanation:

Let the graphic of our function be ${\Psi}_{f}$ with $f \left(x\right) = 4 {x}^{3} - 6 x$, $f : \mathbb{R} \to \mathbb{R}$.

There are two possible situations; the graphic is either symmetric with respect to the axis $x = m$ or to a point $P \left(a , b\right)$.

These two have different definitions, as follow:

A graph ${G}_{f}$ is symmetric to an axis $x = m$ if and only if, from a point on ${G}_{f}$, say $A \left({x}_{A} , {y}_{A}\right)$, the point symmetric to $A \left({x}_{A} , {y}_{A}\right)$ with respect to the axis $x = m$ lies on that graph.

We can see a visual representation of this for parabolas:

graph{(y-x^2)((x-2)^2+(y-4)^2-0.03)((x+2)^2+(y-4)^2-0.03)=0 [-10.005, 9.995, -0.8, 9.2]}

The axis of symmetry for this graph ($y = {x}^{2}$) is $x = 0$.

Basically, in other words, if $x = m$ is an axis of symmetry for the graph of the function $f$, then

$f \left(x\right) = f \left(2 m - x\right)$, $\forall x \in \mathbb{R}$

Before trying it out, let's define what the center of symmetry is.

A point $P \left(a , b\right)$ is the center of symmetry to a graphic ${G}_{f}$ if and only if, for a point $A$ situated on ${G}_{f}$, the point situated the same distance away from point $P$ and on the same line as $A$ and $P$ is also a point of the graph.

We define the mathematical relation so show this the following:

$f \left(2 a - x\right) + f \left(x\right) = 2 b$, $\forall x \in \mathbb{R} \implies f \left(2 a - x\right) = 2 b - f \left(x\right)$

While we could try this out for both relations and do lots of calculations, one simpler way is to obverse that the function $x \mapsto 4 {x}^{3} - 6 x$ is odd, as showed below:

$f \left(x\right) = 4 {x}^{3} - 6 x$
$f \left(- x\right) - 4 {x}^{3} + 6 x$

$\therefore f \left(- x\right) = - f \left(x\right)$

This slightly resembles our second relation, doesn't it? If we put them side to side, we can see the similarity:

$f \left(2 a \textcolor{red}{- x}\right) = 2 b \textcolor{red}{- f \left(x\right)}$
$f \left(\textcolor{R e d}{- x}\right) = \textcolor{red}{- f \left(x\right)}$

In order for these to be the same, we must have $a = b = 0$.

Thus, for an odd function $f$, the graphic of the function ${G}_{f}$ is symmetric to the point $P \left(0 , 0\right)$, which is just the origin, $O$.

Since the function we have, $f \left(x\right) = 4 {x}^{3} - 6 x$, is odd, it means that its graph ${\Psi}_{f}$ is $\textcolor{red}{\text{symmetric with respect to the origin}}$ of the system, $O \left(0 , 0\right)$.

Bonus: If $g$ is an even function, then the graphic of $g$ is symmetric with respect to the axis $x = 0$, which is the $y$-axis.

1

## How do you graph y=-2^x?

Sridhar V.
Featured 2 hours ago

$\text{ }$
Create a data table to graph the exponential function : color(red)(y=-2^x

#### Explanation:

$\text{ }$
color(green)("Step 1 :"

Create a data table:

Closely examine the values for $x$ and $y$

No value of $x$ makes $y$ zero.

So, the graph gets closer and closer to the x-axis but never touches the x-axis.

Hence, color(brown)(y=0 is the horizontal asymptote.

color(green)("Step 2 :"

Using the data table, graph the function $y = f \left(x\right) = - {2}^{x}$

For easy comprehension, graph the exponential function $y = f \left(x\right) = {2}^{x}$, it's parent function.

Compare both the graphs to understand the behavior of the given exponential function $y = f \left(x\right) = - {2}^{x}$

The graph of $y = f \left(x\right) = - {2}^{x}$ is reflected about the x-axis when the sign of $f \left(x\right)$ is negative.

2

## Prove the statement by mathematical induction?

Hammer
Featured 2 hours ago

See below.

#### Explanation:

Since we have two different situations, let's break down this into two different statements, ${\tau}_{1}$ and ${\tau}_{2}$, respectively.

${\tau}_{1} \left(n\right) : 1 + 2 \times {2}^{2} + {3}^{2} + \ldots + 2 \times {\left(n - 1\right)}^{2} + {n}^{2} = {n}^{2} \frac{n + 1}{2}$

if $n$ is odd and

${\tau}_{2} \left(n\right) : 1 + 2 \times {2}^{2} + {3}^{2} + \ldots + {\left(n - 1\right)}^{2} + 2 \times {n}^{2} = n {\left(n + 1\right)}^{2} / 2$

if $n$ is even.

Let's prove ${\tau}_{1}$, firstly. The base case, ${\tau}_{1} \left(1\right)$, claims that

$1 = {1}^{2} \frac{1 + 1}{2} \equiv \text{True}$.

Now, let $x$ be an odd integer such that ${\tau}_{1} \left(x\right)$ is true.

$\therefore \textcolor{red}{1 + 2 \times {2}^{2} + {3}^{2} + \ldots + 2 \times {\left(x - 1\right)}^{2} + {x}^{2} = {x}^{2} \frac{x + 1}{2}} = {S}_{x}$

We must prove if the next case holds. However, the next case is not ${\tau}_{1} \left(x + 1\right)$. This is becaused we defined ${\tau}_{1}$ only on odd positive integers. This means that the next case is actually ${\tau}_{1} \left(x + 2\right)$, as $x + 2$ is odd.

${\tau}_{1} \left(x + 2\right) : \textcolor{red}{1 + 2 \times {2}^{2} + \ldots + 2 \times {\left(x - 1\right)}^{2} + {x}^{2}} + 2 \times {\left(x + 1\right)}^{2} + {\left(x + 2\right)}^{2} = {\left(x + 2\right)}^{2} \frac{x + 3}{2} = {S}_{x + 2}$

Notice how the part highlighted in red is the former sum, ${S}_{x}$.
We know that

${S}_{x} = {x}^{2} \frac{x + 1}{2}$

So by substituting this into the statement, we get

$\textcolor{red}{{x}^{2} \frac{x + 1}{2}} + 2 \times {\left(x + 1\right)}^{2} + {\left(x + 2\right)}^{2} = {\left(x + 2\right)}^{2} \frac{x + 3}{2}$

This is fairly easy to find. Multiply both sides by $2$, expand all squares and multiply the parantheses (or, in other words, do it directly) to get the following result:

${x}^{3} + 7 {x}^{2} + 16 x + 12 = {x}^{3} + 7 {x}^{2} + 16 x + 12$

Which is true. Thus, by mathematical induction, ${\tau}_{1} \left(n\right)$ is true for all odd integers $n$.

The second statement, ${\tau}_{2}$, is similar. Check the base case ${\tau}_{2} \left(2\right)$, assume it is true for some even whole number $y$ then check it for the next even number, $y + 2$.
If you do all that, you will find that ${\tau}_{2} \left(n\right)$ is true for $n$ even.

Hence, by Mathematical induction, the statement

${S}_{n} :$

$\left\{\begin{matrix}1 + 2 \times {2}^{2} + \ldots + 2 \times {\left(n - 1\right)}^{2} + {n}^{2} = {n}^{2} \frac{n + 1}{2} \text{ & if n is odd" \\ 1+2xx2^2+...(n-1)^2+2xxn^2 = n(n+1)^2/2" & if n is even}\end{matrix}\right.$

is true for all $n$ who follow all the respective conditions.