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## The sum of finite series #2/3 + 8/9 + 26/27 +.......to n terms is ? Find the expression.

George C.
Featured 4 months ago

${\sum}_{k = 1}^{n} {a}_{k} = n - \frac{1}{2} + \frac{1}{2} {\left(\frac{1}{3}\right)}^{n}$

#### Explanation:

The general term of a geometric series with initial term $a$ and common ratio $r$ can be written:

${a}_{k} = a \cdot {r}^{k - 1}$

The sum to $n$ terms is given (see footnote) by the formula:

${\sum}_{k = 1}^{n} = \frac{a \left(1 - {r}^{n}\right)}{1 - r}$

Note that:

$\frac{2}{3} = 1 - \frac{1}{3}$

$\frac{8}{9} = 1 - \frac{1}{9}$

$\frac{26}{27} = 1 - \frac{1}{27}$

So it seems that the general term of the given series is:

${a}_{k} = 1 - \frac{1}{3} ^ k$

We have:

${\sum}_{k = 1}^{n} \frac{1}{3} ^ k = \frac{\frac{1}{3} \left(1 - {\left(\frac{1}{3}\right)}^{n}\right)}{1 - \frac{1}{3}}$

$\textcolor{w h i t e}{{\sum}_{k = 1}^{n} \frac{1}{3} ^ k} = \frac{1 - {\left(\frac{1}{3}\right)}^{n}}{2}$

$\textcolor{w h i t e}{{\sum}_{k = 1}^{n} \frac{1}{3} ^ k} = \frac{1}{2} - \frac{1}{2} {\left(\frac{1}{3}\right)}^{n}$

So:

${\sum}_{k = 1}^{n} {a}_{k} = {\sum}_{k = 1}^{n} \left(1 - \frac{1}{3} ^ k\right)$

$\textcolor{w h i t e}{{\sum}_{k = 1}^{n} {a}_{k}} = {\sum}_{k = 1}^{n} 1 - {\sum}_{k = 1}^{n} \frac{1}{3} ^ k$

$\textcolor{w h i t e}{{\sum}_{k = 1}^{n} {a}_{k}} = n - \left(\frac{1}{2} - \frac{1}{2} {\left(\frac{1}{3}\right)}^{n}\right)$

$\textcolor{w h i t e}{{\sum}_{k = 1}^{n} {a}_{k}} = n - \frac{1}{2} + \frac{1}{2} {\left(\frac{1}{3}\right)}^{n}$

$\textcolor{w h i t e}{}$
Footnote

Given:

${a}_{k} = a {r}^{k - 1}$

We find:

$\left(1 - r\right) {\sum}_{k = 1}^{n} {a}_{k} = \left(1 - r\right) {\sum}_{k = 1}^{n} a {r}^{k - 1}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{k = 1}^{n} {a}_{k}} = {\sum}_{k = 1}^{n} a {r}^{k - 1} - r {\sum}_{k = 1}^{n} a {r}^{k - 1}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{k = 1}^{n} {a}_{k}} = {\sum}_{k = 1}^{n} a {r}^{k - 1} - {\sum}_{k = 2}^{n + 1} a {r}^{k - 1}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{k = 1}^{n} {a}_{k}} = a + \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{k = 2}^{n} a {r}^{k - 1}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{k = 2}^{n} a {r}^{k - 1}}}} - a {r}^{n}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{k = 1}^{n} {a}_{k}} = a \left(1 - {r}^{n}\right)$

So dividing both ends by $\left(1 - r\right)$ we get:

${\sum}_{k = 1}^{n} {a}_{k} = \frac{a \left(1 - {r}^{n}\right)}{1 - r}$

Cesareo R.
Featured 3 months ago

See below.

#### Explanation:

5)
Let $d \left(n\right)$ be the function which counts the number of digits of $n$ then

5.1) $n \ge {10}^{d \left(n\right)} \ge g \left(n\right)$
5.2) ${\left(n - 6\right)}^{2} = g \left(n\right) \to n = \sqrt{g \left(n\right)} + 6$ then
$g \left({n}_{k}\right) = {k}^{2} , k = 1 , 2 , 3 , \cdots$ so the equation to obey is

${n}_{k} = k + 6$ so the only number observing this relationship is

${n}_{1} = 7$

6)
Let $4 + {p}_{1} {p}_{2} = {m}_{1}^{2}$ and $4 + {p}_{1} {p}_{3} = {m}_{2}^{2}$
then

$\left\{\begin{matrix}{p}_{1} {p}_{2} = \left({m}_{1} - 2\right) \left({m}_{1} + 2\right) \\ {p}_{1} {p}_{3} = \left({m}_{2} - 2\right) \left({m}_{2} + 2\right)\end{matrix}\right.$

but ${p}_{1} , {p}_{2} , {p}_{3}$ being prime numbers then

$\left\{\begin{matrix}{p}_{1} = {m}_{1} - 2 \\ {p}_{2} = {m}_{1} + 2 \\ {p}_{1} = {m}_{2} + 2 \text{ otherwise } {m}_{1} = {m}_{2} \\ {p}_{3} = {m}_{2} - 2\end{matrix}\right.$

so

${m}_{1} = {m}_{2} + 4$ then

$\left\{\begin{matrix}{p}_{1} = {p}_{3} + 4 \\ {p}_{2} = {p}_{1} + 4\end{matrix}\right.$

and

${p}_{2} > {p}_{1} > {p}_{3}$

so the prime numbers are

${p}_{1} - 4 , {p}_{1} , {p}_{1} + 4$

or

${p}_{3} = 3 , {p}_{1} = 7 , {p}_{2} = 11$

## How are the logs coming about? thanks

George C.
Featured 2 months ago

A few thoughts...

#### Explanation:

I am not exactly sure what you are asking, but here's something of what logarithms are about...

Let $b > 1$ be a constant.

Consider the function:

$f \left(x\right) = {b}^{x}$

This is a continuous, one to one, strictly monotonically increasing function from $\left(- \infty , \infty\right)$ onto $\left(0 , \infty\right)$ looking something like this:

graph{2^x [-9.58, 10.42, -2.4, 7.6]}

It has the interesting property that:

$f \left(x + y\right) = {b}^{x + y} = {b}^{x} {b}^{y} = f \left(x\right) f \left(y\right)$

for any $x , y \in \left(- \infty , \infty\right)$

The inverse of $f \left(x\right)$ has a graph like this:

graph{2^y-x = 0 [-4.08, 15.92, -4.12, 5.88]}

which is a continuous, one to one, strictly monotonically increasing function from $\left(0 , \infty\right)$ onto $\left(- \infty , \infty\right)$

Given any $u , v \in \left(0 , \infty\right)$, let:

$x = {f}^{- 1} \left(u\right)$

$y = {f}^{- 1} \left(v\right)$

Then we find:

${f}^{- 1} \left(u v\right) = {f}^{- 1} \left(f \left(x\right) f \left(y\right)\right) = {f}^{- 1} \left(f \left(x + y\right)\right) = x + y = {f}^{- 1} \left(u\right) + {f}^{- 1} \left(v\right)$

This inverse function ${f}^{- 1} \left(x\right)$ is the logarithm base $b$

and this property we have found is:

${\log}_{b} \left(u v\right) = {\log}_{b} \left(u\right) + {\log}_{b} \left(v\right)$

a fundamentally useful property of logarithms.

This property of logarithms was discovered by John Napier in the 17th century and used in the form of slide rules and tables to help perform calculations.

## Can someone prove to me that multiplication by imaginary numbers is a rotation?

George C.
Featured 1 month ago

A few thoughts...

#### Explanation:

The Real numbers are usually thought of as constituting a line which we call the Real line.

This is the $x$-axis of the Complex plane, representing an extension of the Real numbers to numbers of the form $a + b i$, where $i$ is the imaginary unit (i.e. the point $\left(0 , 1\right)$).

$\left(a + b i\right) + \left(c + \mathrm{di}\right) = \left(a + c\right) + \left(b + d\right) i$

Multiplication of complex numbers is defined as:

$\left(a + b i\right) \left(c + \mathrm{di}\right) = \left(a c - b d\right) + \left(a d + b c\right) i$

This has some interesting properties:

• Real numbers are complex numbers of the form $a + 0 i$

• Multiplication by Real numbers is scalar multiplication:

$\left(a + 0 i\right) \left(c + \mathrm{di}\right) = a c + a \mathrm{di}$

• The square of $i$ is $- 1$:

${i}^{2} = \left(0 + 1 i\right) \left(0 + 1 i\right) = \left(\left(0\right) \left(0\right) - \left(1\right) \left(1\right)\right) + \left(\left(0\right) \left(1\right) + \left(1\right) \left(0\right)\right) i = - 1$

The modulus of a complex number is its length as a vector, which can be deduce from Pythagoras:

$\left\mid a + b i \right\mid = \sqrt{{a}^{2} + {b}^{2}}$

Hence we find that any Complex number of modulus $1$ can be represented in the form:

$\cos \theta + i \sin \theta$

Multiplication by such a number is pure rotation by $\theta$ in an anticlockwise direction:

$\left(\cos \theta + i \sin \theta\right) \left(a + b i\right)$

$= \left(a \cos \theta - b \sin \theta\right) + \left(b \cos \theta + a \sin \theta\right) i$

By the time we get to this point, some readers might be thinking "matrices". Indeed, complex numbers have a natural representation in terms of $2 \times 2$ matrices with real coefficients. Then addition and multiplication become matrix addition and multiplication:

$\left(\begin{matrix}a & b \\ - b & a\end{matrix}\right) + \left(\begin{matrix}c & d \\ - d & c\end{matrix}\right) = \left(\begin{matrix}a + c & b + d \\ - \left(b + d\right) & a + c\end{matrix}\right)$

$\left(\begin{matrix}a & b \\ - b & a\end{matrix}\right) \left(\begin{matrix}c & d \\ - d & c\end{matrix}\right) = \left(\begin{matrix}a c - b d & a d + b c \\ - \left(a d + b c\right) & a c - b d\end{matrix}\right)$

Notice that complex numbers of modulus $1$ take the form:

$\left(\begin{matrix}\cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{matrix}\right)$

which is recognisable as the matrix representing a rotation through angle $\theta$ about the origin.

In general, multiplication by a complex number is a combination of rotation about the origin and scaling.

## How would you graph #y= -lnx# ?

George C.
Featured 2 weeks ago

Take the graph of $y = {e}^{x}$, reflect it in the line $y = x$, then in the $x$ axis.

#### Explanation:

Do you know what the graph of $y = {e}^{x}$ looks like?

graph{y=e^x [-10, 10, -5, 5]}

• It is monotonically increasing.
• It is always greater than $0$, so lies completely above the $x$ axis.
• It is rapidly asymptotic to the $x$ axis for negative values of $x$.
• It intersects the $y$ axis at $\left(0 , 1\right)$.
• It grows very rapidly for positive values of $x$.

Next note that $\ln x$ is the inverse function of ${e}^{x}$.

So the graph of $y = \ln x$ can be found by swapping $x$ and $y$, that is by reflecting the above graph in the diagonal line $y = x$, to get:

graph{y=ln x [-10, 10, -5, 5]}

Note that:

• It is monotonically increasing.
• It is only defined for $x > 0$, so the graph lies entirely to the right of the $y$ axis.
• It has a vertical asymptote at $x = 0$.
• It intersects the $x$ axis at $\left(1 , 0\right)$.
• It grows very slowly for positive values of $x$.

Finally, to get the graph of $y = - \ln x$ we just have to reflect the above graph in the $x$ axis to get:

graph{y=-ln x [-10, 10, -5, 5]}

Note that:

• It is monotonically decreasing.
• It is only defined for $x > 0$, so the graph lies entirely to the right of the $y$ axis.
• It has a vertical asymptote at $x = 0$.
• It intersects the $x$ axis at $\left(1 , 0\right)$.
• It grows more negative very slowly for positive values of $x$.

## How do you use synthetic substitution to find f(3) for #f(x)=x^2 -9x+ 5#?

Alan P.
Featured 6 days ago

$f \left(\textcolor{g r e e n}{3}\right) = \textcolor{red}{- 13}$

#### Explanation:

(Sorry, but I had to do this as a graphic image to get the arrows in.)

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