heureka

I assume: Compute 1*1/2+2*1/4+3*1/8+...+n*1/2^n+...

Compute:

\(\displaystyle s = 1\cdot \dfrac12+2\cdot\dfrac14+ 3\cdot \dfrac18+ 4\cdot \dfrac{1}{16}+\ldots + n\cdot \dfrac{1}{2^n}+\ldots \infty\)

\(\begin{array}{|lrll|} \hline &s =& \displaystyle 1\cdot \dfrac12+2\cdot\dfrac14+ 3\cdot \dfrac18+ 4\cdot \dfrac{1}{16}+\ldots + n\cdot \dfrac{1}{2^n}+\ldots \infty \\\\ &s =& \displaystyle 1\cdot \left(\dfrac12\right)^1+2\cdot\left(\dfrac12\right)^2+ 3\cdot \left(\dfrac12\right)^3+ 4\cdot \left(\dfrac{1}{2}\right)^4+\ldots + n\cdot \left(\dfrac{1}{2} \right)^n+\ldots \infty \\\\ && \displaystyle\qquad \text{We substitute:} \quad \boxed{x=\dfrac12} \\\\ &s =& \displaystyle 1x^1+2x^2+ 3x^3+ 4x^4+\ldots + nx^n+\ldots \infty \quad | \quad : x \\\\ &\dfrac{s}{x} =& \displaystyle 1x^0+2x^1+ 3x^2+ 4x^3+\ldots + nx^{n-1}+\ldots \infty \\\\ &\dfrac{s}{x} =& \displaystyle \sum \limits_{n=1}^{\infty} nx^{n-1} \quad | \quad \cdot \ dx \\\\ &\dfrac{s}{x}\ dx =& \displaystyle \sum \limits_{n=1}^{\infty} \left( nx^{n-1}\ dx \right) \quad | \quad \text{integrate both sides} \\\\ &\int \dfrac{s}{x}\ dx =& \displaystyle \int \left(\sum \limits_{n=1}^{\infty} \left( nx^{n-1}\ dx \right) \right) \\\\ &\int \dfrac{s}{x}\ dx =& \displaystyle \sum \limits_{n=1}^{\infty} x^{n} \quad | \quad \text{This is the sum of a infinite geometric series},\ a_1=x,\ r=x \\\\ && \displaystyle\qquad \text{The sum of this infinite geometric series is:} \quad \boxed{= \dfrac{x}{1-x},\ \quad |x| \lt 1 } \\\\ &\int \dfrac{s}{x}\ dx =& \displaystyle \dfrac{x}{1-x} \quad | \quad \text{derivate both sides} \\\\ & \dfrac{s}{x} =& \displaystyle 1\cdot(1-x)^{-1}+x\cdot(-1)\cdot(1-x)^{-2}\cdot(-1) \\\\ & \dfrac{s}{x} =& \displaystyle \dfrac{1}{1-x} + \dfrac{x}{(1-x)^2} \\\\ & \dfrac{s}{x} =& \displaystyle \dfrac{1-x+x}{(1-x)^2} \\\\ & \dfrac{s}{x} =& \displaystyle \dfrac{1}{(1-x)^2} \\\\ & \mathbf{s=}& \mathbf{\displaystyle \dfrac{x}{(1-x)^2}} \quad | \quad x=\dfrac12 \\\\ & s= & \displaystyle \dfrac{\dfrac12}{\left(1-\dfrac12 \right)^2} \\\\ & s= & \displaystyle \dfrac{\dfrac12}{\left(\dfrac12 \right)^2} \\\\ & s= &\displaystyle \dfrac{1}{ \dfrac12 }\\\\ & \mathbf{s=}& \mathbf{\displaystyle 2} \\ \hline \end{array}\)

\(\displaystyle 1\cdot \dfrac12+2\cdot\dfrac14+ 3\cdot \dfrac18+ 4\cdot \dfrac{1}{16}+\ldots + n\cdot \dfrac{1}{2^n}+\ldots \infty = 2\)

a)

(4x) * (1+2x^2)^-0,5 dx Grenzen 2 und 0 durch Substitution.

Integral

\(\displaystyle \int \limits_0^{2}\dfrac{4x}{\sqrt{1+2x^2}}\ dx = \ ?\)

\(\begin{array}{|lrcl|} \hline \displaystyle \int \limits_0^{2}\dfrac{4x}{\sqrt{1+2x^2}}\ dx =\ ? \\\\ & \text{Substitution:}\\ & \mathbf{z = \sqrt{1+2x^2}} &=& (1+2x^2)^{\frac12} \\ & dz &=& \frac12 \cdot (1+2x^2)^{\frac12 - 1} \cdot 2\cdot 2x \ dx \\ & dz &=& 2x \cdot (1+2x^2)^{-\frac12} \ dx \\ & dz &=& \dfrac{ 2x } { \sqrt{1+2x^2} } \ dx \\ & \mathbf{dx} &\mathbf{=}& \mathbf{\dfrac{ \sqrt{1+2x^2} } { 2x } \ dz} \\\\ & \text{Neue Grenzen:}\\ & x&=&0 \quad \Rightarrow \quad z = \sqrt{1+2\cdot 0^2} = 1 \\ & x&=&2 \quad \Rightarrow \quad z = \sqrt{1+2\cdot 2^2} = 3 \\ \hline \end{array}\)

\(\begin{array}{|rcl|} \hline \displaystyle \mathbf{\int \limits_0^{2} \dfrac{4x}{\sqrt{1+2x^2}}\ dx } \\\\ &=& \displaystyle \int \limits_{z=1}^{z=3}\dfrac{4x}{\sqrt{1+2x^2}}\cdot \dfrac{ \sqrt{1+2x^2} } { 2x } \ dz \\\\ &=& \displaystyle \int \limits_{z=1}^{z=3} 2 \ dz \\\\ &=& \displaystyle 2\int \limits_{z=1}^{z=3} \ dz \\\\ &=& \displaystyle 2[z]_{z=1}^{z=3} \\\\ &=& \displaystyle 2[3-1] \\\\ &=& \displaystyle 2\cdot 2 \\\\ \mathbf{\displaystyle \int \limits_0^{2}\dfrac{4x}{\sqrt{1+2x^2}}\ dx }& \mathbf{=} & \mathbf{4} \\ \hline \end{array}\)

Find the constant term in the expansion of $$\left(10x^3-\frac{1}{2x^2}\right)^{5}$$

\(\begin{array}{|rcll|} \hline && \left(10x^3-\dfrac{1}{2x^2}\right)^{5} \\\\ &=& \left(\dfrac{10x^3\cdot 2x^2-1}{2x^2}\right)^{5} \\\\ &=& \left(\dfrac{20x^5-1}{2x^2}\right)^{5} \\\\ &=&\dfrac{1}{\left(2x^2 \right)^5} \left( 20x^5-1 \right)^{5} \\\\ &=&\dfrac{1}{ 2^5x^{10} } \left( 20x^5-1 \right)^{5} \\\\ &=&\dfrac{1}{ 32x^{10} } \left( 20x^5-1 \right)^{5} \\\\ &=&\dfrac{1}{ 32x^{10} } \left[ \dbinom{5}{0}\left(20x^5 \right)^5 -\dbinom{5}{1}\left(20x^5 \right)^4 +\dbinom{5}{2}\left(20x^5 \right)^3 \\ -\dbinom{5}{3}\left(20x^5 \right)^2 +\dbinom{5}{4}\left(20x^5 \right) +\dbinom{5}{5}(-1)^5 \right] \\\\ &=&\dfrac{1}{ 32x^{10} } \left[ \dbinom{5}{0} 20^5x^{25} -\dbinom{5}{1} 20^4x^{20} +\dbinom{5}{2} 20^3x^{15} \\ { \color{red}-\dbinom{5}{3} 20^2x^{10} } +\dbinom{5}{4} 20x^5 -\dbinom{5}{5} \right] \\ \hline \end{array}\)

Constant term:

\(\begin{array}{|rcll|} \hline && \dfrac{1}{ 32x^{10} }\left[ { \color{red}-\dbinom{5}{3} 20^2x^{10} } \right] \\\\ &=& -\dfrac{1}{ 32x^{10} } \dbinom{5}{3} 20^2x^{10} \\\\ &=& -\dfrac{20^2}{ 32 } \dbinom{5}{3} \quad & | \quad \dbinom{5}{3} = \dbinom{5}{5-3}=\dbinom{5}{2}=\dfrac{5}{2}\cdot \dfrac{4}{1} = 10 \\\\ &=& -\dfrac{20^2\cdot 10}{ 32 } \\\\ &=& -\dfrac{4000}{ 32 } \\\\ &\mathbf{=}& \mathbf{-125} \\ \hline \end{array}\)

Find all numbers r for which the system of congruences:

\(\begin{align*} x &\equiv r \pmod{6}, \\ x &\equiv 9 \pmod{20}, \\ x &\equiv 4 \pmod{45} \end{align*}\)

has a solution.

\(\begin{array}{|rclll|} \hline x &\equiv& 9 \pmod{20} \\ x &\equiv& 4 \pmod{45} \\ \text{or} \\ x &=& 9 + 20n \quad | \quad n \in Z \\ x &=& 4 + 45m \quad | \quad m \in Z \\\\ && & x = 9 + 20n = 4 + 45m \\\\ && & 20n = 45m - 5 \quad & | \quad : 5 \\\\ && & 4n = 9m - 1 \\\\ && & \mathbf{ n = \dfrac{9m-1}{4} }\\\\ && & n = \dfrac{8m+m-1}{4} \\\\ && & n = 2m+\underbrace{\dfrac{m-1}{4}}_{=a} \\\\ && && a = \dfrac{m-1}{4} \\\\ && && 4a = m-1 \\\\ && & & \mathbf{ m = 4a + 1 } \\\\ && & \mathbf{ n = \dfrac{9(4a + 1 )-1}{4} }\\\\ && & n = 9a + \dfrac{8}{4} \\\\ && & \mathbf{ n = 9a + 2 }\\\\ \mathbf{x} &\mathbf{=}& \mathbf{9 + 20(9a + 2)} \\\\ x & = & 9+180a+40 \\\\ \mathbf{ x } &\mathbf{ = } & \mathbf{49+180a} \quad | \quad n \in Z \\ \hline \end{array} \)

\(\begin{align*} x &\pmod{20}:& &49\pmod{20}+(180a)\pmod{20} \\ && &= 9 + 0 \\ && &= 9\ \checkmark \\\\ x &\pmod{45}:& &49\pmod{45}+(180a)\pmod{45} \\ && &= 4 + 0 \\ && &= 4\ \checkmark \\\\ r &\equiv x \pmod{6}:&&r = 49\pmod{6}+(180a)\pmod{6} \\ && &r = 1 + 0 \\ && & \mathbf{r = 1} \\ \end{align*}\)

What is 3/2 to the 3rd power?

\(\begin{array}{|rcll|} \hline && \mathbf{ \left(\dfrac32 \right)^3 } \\\\ &=& \left(\dfrac32 \right)\cdot \left(\dfrac32 \right)\cdot \left(\dfrac32 \right) \\\\ &=& \dfrac{3\cdot 3\cdot 3 } {2\cdot 2 \cdot 2} \\\\ &=& \dfrac{27} {8} \\\\ &\mathbf{=}& \mathbf{3.375} \\ \hline \end{array}\)

Thank you, Guest !

What is the largest negative integer \(x\) satisfying \(24x \equiv 15 \pmod{1199}~?\)

\(\begin{array}{|rclll|} \hline 24x &\equiv& 15 \pmod{1199} \\ \text{or} \\ 24x &=& 15 + 1199k \quad & | \quad k \in Z \\\\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{15 + 1199k}{24}} \\\\ x &=& \dfrac{49\cdot 24k + 23k + 15 }{24} \\\\ x &=& 49k + \underbrace{ \dfrac{23k + 15 }{24} }_{=a} \\\\ && & a = \dfrac{23k + 15 }{24} \\\\ && & 24a = 23k + 15 \\\\ && & 23k = 24a - 15 \\\\ && & \mathbf{ k = \dfrac{24a - 15}{23} }\\\\ && & k = \dfrac{23a+a - 15}{23} \\\\ && & k = a + \underbrace{ \dfrac{a - 15}{23} }_{=b} \\\\ && && b = \dfrac{a - 15}{23} \\\\ && && 23b = a - 15 \\\\ && & & \mathbf{ a = 23b + 15} \\\\ && & \mathbf{ k = \dfrac{24(23b + 15) - 15}{23} }\\\\ && & k = \dfrac{24\cdot 23b + 24\cdot 15 - 15}{23} \\\\ && & k = \dfrac{24\cdot 23b + 23\cdot 15}{23} \\\\ && & \mathbf{k = 24b+15} \\\\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{15 + 1199(24b+15)}{24}} \\\\ x & = & \dfrac{15 + 1199\cdot 24b+ 1199\cdot 15}{24} \\\\ x & = & \dfrac{1199\cdot 24b+ 1200\cdot 15}{24} \\\\ x & = & \dfrac{1199\cdot 24b+ 50\cdot 24 \cdot 15}{24} \\\\ x & = & 50\cdot 15 + 1199 b \\\\ x & = & 750 + 1199 b \quad & | \quad b = -1 \\\\ x & = & 750 - 1199 \\\\ \mathbf{ x } &\mathbf{ = } & \mathbf{-449} \\ \hline \end{array}\)

Calculate

Calculate:

\(\begin{array}{|lcll|} \hline \mathbf{ S_n = \dfrac{1}{1 \cdot 2 \cdot 3} + \dfrac{1}{2 \cdot 3 \cdot 4} + \dfrac{1}{3 \cdot 4 \cdot 5} + \dfrac{1}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{1}{n \cdot (n+1) \cdot (n+2)}} \\ \hline \end{array}\)

Formula:

\(\begin{array}{|lcll|} \hline \text{in general}:\ \dfrac{1}{n(n+d)} = \dfrac{1}{d}\left(\dfrac{1}{n}- \dfrac{1}{n+d} \right) \\ \hline \\ \begin{array}{lrcll} \text{we need}: & \dfrac{1}{(n+1)(n+2)} &=& \dfrac{1}{n+1}-\dfrac{1}{n+2} \\\\ & \dfrac{1}{n(n+1)} &=& \dfrac{1}{n}-\dfrac{1}{n+1} \\\\ & \dfrac{1}{n(n+2)} &=& \dfrac{1}{2} \left( \dfrac{1}{n}-\dfrac{1}{n+2} \right) \\ \end{array} \\ \hline \end{array}\)

We rearrange:

\(\begin{array}{|rcll|} \hline \dfrac{1}{n \cdot (n+1) \cdot (n+2)} \\\\ &=& \dfrac{1}{n}\times \dfrac{1}{(n+1) \cdot (n+2)} \\\\ &=& \dfrac{1}{n}\times \left( \dfrac{1}{n+1}-\dfrac{1}{n+2} \right) \\\\ &=& \dfrac{1}{n}\times \dfrac{1}{n+1} - \dfrac{1}{n}\times \dfrac{1}{n+2} \\\\ &=& \left(\dfrac{1}{n}-\dfrac{1}{n+1} \right)- \dfrac{1}{2} \times \left(\dfrac{1}{n} -\dfrac{1}{n+2} \right) \\\\ &=& \dfrac{1}{n} - \dfrac{1}{n+1} -\dfrac{1}{2n} + \dfrac{1}{2(n+2)} \\\\ \mathbf{\dfrac{1}{n \cdot (n+1) \cdot (n+2)} } & \mathbf{=} & \mathbf{ \dfrac{1}{2n} - \dfrac{1}{n+1} + \dfrac{1}{2(n+2)} } \\ \hline \end{array}\)

Telescoping series:

\(\begin{array}{|rcll|} \hline S_n &=& \mathbf{\dfrac{1}{2}} &\mathbf{-}& \mathbf{\dfrac{1}{2}} &\color{red}+& \color{red}\dfrac{1}{6} \\\\ &\mathbf{+}& \mathbf{\dfrac{1}{4}} &\color{red}-& \color{red}\dfrac{1}{3} &\color{blue}+& \color{blue}\dfrac{1}{8} \\\\ &\color{red}+& \color{red}\dfrac{1}{6} &\color{blue}-& \color{blue}\dfrac{1}{4} &\color{red}+& \color{red}\dfrac{1}{10} \\\\ &\color{blue}+& \color{blue}\dfrac{1}{8} &\color{red}-& \color{red}\dfrac{1}{5} &\color{green}+& \color{green}\dfrac{1}{12} \\\\ && \ldots \\\\ &+\color{red}& \color{red}\dfrac{1}{2(n-2)} &\color{green}-& \color{green}\dfrac{1}{n-1} &\color{red}+& \color{red}\dfrac{1}{2n} \\\\ &\color{green}+& \color{green}\dfrac{1}{2(n-1)} &\color{red}-& \color{red}\dfrac{1}{n} &\mathbf{+}& \mathbf{\dfrac{1}{2(n+1)}} \\\\ &\color{red}+& \color{red}\dfrac{1}{2n} &\mathbf{-}& \mathbf{\dfrac{1}{n+1}} &\mathbf{+}& \mathbf{\dfrac{1}{2(n+2)}} \\ \hline \end{array}\)

The part of each term cancelling with part of the next two diagonal terms:
Example:

\(\begin{array}{|lcll|} \hline \dfrac{1}{6}-\dfrac{1}{3}+\dfrac{1}{6} = 0 \\\\ \dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{8} = 0 \\\\ \dfrac{1}{10}-\dfrac{1}{5}+\dfrac{1}{10} = 0 \\\\ \ldots \\\\ \dfrac{1}{2n}-\dfrac{1}{n} + \dfrac{1}{2n} = 0 \\ \hline \end{array} \)

So \(S_n\) is, we have all black terms left :

\(\begin{array}{|rcll|} \hline S_n &=& \dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{4} + \dfrac{1}{2(n+1)} - \dfrac{1}{n+1} + \dfrac{1}{2(n+2)} \\\\ &=& \dfrac{1}{4} - \dfrac{1}{2(n+1)} + \dfrac{1}{2(n+2)} \\\\ &=& \dfrac{1}{4} - \dfrac{1}{2}\left( \dfrac{1}{n+1} - \dfrac{1}{n+2} \right) \\\\ &=& \dfrac{1}{4} - \dfrac{1}{2}\left( \dfrac{1}{(n+1)(n+2)} \right) \\\\ &=& \dfrac{1}{4} - \dfrac{1}{2(n+1)(n+2)} \\\\ &=& \dfrac{1}{4} \left(1- \dfrac{2}{(n+1)(n+2)} \right) \\\\ &=& \dfrac{(n+1)(n+2)-2}{4(n+1)(n+2)} \\\\ &=& \dfrac{n(n+2)+n+2-2}{4(n+1)(n+2)} \\\\ &=& \dfrac{n(n+2)+n}{4(n+1)(n+2)} \\\\ &=& \dfrac{n(n+2+1)}{4(n+1)(n+2)} \\\\ \mathbf{ S_n }& \mathbf{=} & \mathbf{ \dfrac{n(n+3)}{4(n+1)(n+2)} } \\ \hline \end{array}\)

\(\large{ \mathbf{ S_n = \sum \limits_{k=1 }^{n} \dfrac 1{k(k+1)(k+2)} = \dfrac{n(n+3)}{4(n+1)(n+2)} } }\)

I'm having a rough time figuring out the next 3 terms of this sequence.
1/2,  1/3,  1/6,  1/15,  1/45,  2/315.......

\(\begin{array}{|rcll|} \hline \dfrac12 \times \boxed{\dfrac{2}{3}} &=& \dfrac13 \\\\ \dfrac13 \times \boxed{\dfrac{2}{4}} &=& \dfrac16 \\\\ \dfrac16 \times \boxed{\dfrac{2}{5}} &=& \dfrac{1}{15} \\\\ \dfrac{1}{15} \times \boxed{\dfrac{2}{6}} &=& \dfrac{1}{45} \\\\ \dfrac{1}{45} \times \boxed{\dfrac{2}{7}} &=& \dfrac{2}{315} \\\\ \dfrac{2}{315} \times \boxed{\dfrac{2}{8}} &=& \dfrac{1}{630} \\\\ \dfrac{1}{630} \times \boxed{\dfrac{2}{9}} &=& \dfrac{1}{2835} \\\\ \dfrac{1}{2835} \times \boxed{\dfrac{2}{10}} &=& \dfrac{1}{14175} \\ \hline \end{array}\)

3)
Sumalee's school is selling tickets to a spring musical.
On the first day of ticket  sales the school sold 6 adult tickets and 1 student ticket for a total of $23.
The school took in $106 on the second day by selling 12 adult tickets and 14 student tickets.
Find the price of an adult ticket and the price of a student ticket.

x = adult ticket price
y = student ticket price

\(\begin{array}{|lrcll|} \hline (1) & 6x+1y &=& $23 \\ & y &=& 23 - 6x \\\\ (2) & 12x+14y &=& $106 \\ & 12x+14(23 - 6x) &=& 106 \quad &| \quad :2 \\ & 6x+7(23 - 6x) &=& 53 \\ & 6x +161 - 42x &=& 53 \\ & 161 - 36x &=& 53 \quad &| \quad + 36x \\ & 161 &=& 53 + 36x \quad &| \quad - 53 \\ & 108 &=& 36x \quad &| \quad : 36 \\ & 3 &=& x \\ & \mathbf{x} & \mathbf{=} & \mathbf{$3} \\\\ & y &=& 23 - 6x \quad &| \quad x=3 \\ & y &=& 23 - 6(3) \\ & y &=& 23 - 18 \\ & y &=& 5 \\ & \mathbf{y} & \mathbf{=} & \mathbf{$5} \\ \hline \end{array} \)

The price of a adult ticket is $3.
The price of a student ticket is $5.