heureka

make "p" the subject.
\(\displaystyle x={5 \over3}- \sqrt{2p-1}\)

\(\begin{array}{|rcll|} \hline x &=& \dfrac{5}{3} - \sqrt{2p-1} \quad & | \quad -\dfrac{5}{3} \\ x-\dfrac{5}{3} &=& - \sqrt{2p-1} \quad & | \quad \times(-1) \\ -x+\dfrac{5}{3} &=& \sqrt{2p-1} \\ \dfrac{5}{3}-x &=& \sqrt{2p-1} \quad & | \quad \text{square both sides} \\ \left(\dfrac{5}{3}-x\right)^2 &=& 2p-1 \quad & | \quad +1 \\ \left(\dfrac{5}{3}-x\right)^2+1 &=& 2p \quad & | \quad :2 \\ \mathbf{\dfrac{ \left(\dfrac{5}{3}-x\right)^2+1 } {2}} & \mathbf{=}& \mathbf{p} \\ \hline \end{array}\)

Determine, without a calculator, which is greater 4^100 or 6^75

\(\begin{array}{|rcl|cl|} \hline && 4^{100} && 6^{75} \\ &=& 4^{4\cdot 25} &=& 6^{3\cdot 25} \\ &=& \left(4^4\right)^{25} &=& \left(6^3\right)^{25} \\ &=& 256^{25} &=& 216^{25} \\ \hline \end{array} \)

\(\text{So $4^{100}$ is greater than $6^{75}$ }\)

Wie kann ich die Formel:
\(\displaystyle \large{q^1 + q^2 + q^3 +\ldots+ q^n = \dfrac{q^{n+1} - q}{q-1} }\)
durch völlstandige Induktion beweisen ?
für alle \( n \in \mathbb{N}\):

\(\bf{\text{Beweise mit vollständiger Induktion:}} \\ \displaystyle q^1 + q^2 + q^3 +\ldots+ q^n = \dfrac{q^{n+1} - q}{q-1} \)

\(\text{Induktionsanfang:}\)

\(\begin{array}{|lll|} \hline n=1 & \text{linke Seite:} & q^1 \\ & & =q \\\\ & \text{rechte Seite:} & \dfrac{q^{1+1} - q} {q-1} \\ & &= \dfrac{q^{2} - q} {q-1} \\ & &= \dfrac{q\cdot q - q} {q-1} \\ & &= \dfrac{q\cdot( q - 1)} {q-1} \\ & &= q\cdot \left( \dfrac{q - 1} {q-1} \right) \\ & &= q \\ \hline \end{array}\)

\(\text{Für $\mathbf{n=1}$ sind beide Seiten gleich, und die Aussage ist wahr!}\)

\(\text{Die Induktionsannahme (I.A.) lautet:}\)

\(\begin{array}{|rcll|} \hline \displaystyle q^1 + q^2 + q^3 +\ldots+ q^n = \dfrac{q^{n+1} - q}{q-1} \\ \hline \end{array}\)

\(\text{Der Induktionsschluss von $\mathbf{n}$ nach $\mathbf{n+1}$:}\)

\(\begin{array}{|rcll|} \hline \displaystyle q^1 + q^2 + q^3 +\ldots+ q^n + q^{n+1} &=& \dfrac{q^{(n+1)+1} - q}{q-1} \\ \hline \end{array}\)

\(\bf{\text{linke Seite:}}\)

\(\begin{array}{|llrcll|} \hline &\mathbf{ q^1 + q^2 + q^3 +\ldots+ q^n + q^{n+1} }\\\\ \overset{I.A.}{=} & \dfrac{q^{n+1} - q}{q-1} + q^{n+1} \\\\ = & \dfrac{q^{n+1} - q}{q-1} + q^{n+1}\cdot \left(\dfrac{q-1}{q-1}\right) \\\\ = & \dfrac{q^{n+1} - q+ q^{n+1}(q-1)}{q-1} \\\\ = & \dfrac{q^{n+1} - q+ q^{n+2}-q^{n+1} }{q-1} \\\\ = & \dfrac{ - q+ q^{n+2}}{q-1} \\\\ \mathbf{=} & \mathbf{\dfrac{ q^{n+2}- q }{q-1} }\\ \hline \end{array} \)

\(\bf{\text{rechte Seite:}}\)

\(\begin{array}{|ll|} \hline & \mathbf{\dfrac{q^{(n+1)+1} - q}{q-1} } \\\\ \mathbf{=}& \mathbf{\dfrac{ q^{n+2}- q }{q-1} }\\ \hline \end{array}\)

Bekah has four brass house number digits:
2, 3, 5 and 7, and only has one of each number.
How many distinct numbers can she form using one or more of the digits?

\(\text{binary code} \\ \begin{array}{|c|c|c|c|r|c|} \hline 2 & 3 & 5 & 7 & & \text{distinct numbers} \\ \hline 0&0&0&1 & 7 & 1! \\ 0&0&1&0 & 5 & 1! \\ 0&0&1&1 & 57 \text{ or } 75 & 2! \\ 0&1&0&0 & 3 & 1! \\ 0&1&0&1 & 37 \text{ or } 73 & 2! \\ 0&1&1&0 & 35 \text{ or } 53 & 2! \\ 0&1&1&1 & 357 & 3! \\ 1&0&0&0 & 2 & 1! \\ 1&0&0&1 & 27 \text{ or } 72 & 2! \\ 1&0&1&0 & 25 \text{ or } 52 & 2! \\ 1&0&1&1 & 257 & 3! \\ 1&1&0&0 & 23 \text{ or } 32 & 2! \\ 1&1&0&1 & 237 & 3! \\ 1&1&1&0 & 235 & 3! \\ 1&1&1&1 & 2357 & 4! \\ \hline &&&&\text{sum}& = 4\times 1! + 6\times 2! + 4\times 3!+1\times 4! \\ &&&&& = \dbinom{4}{1}1! +\dbinom{4}{2}2! +\dbinom{4}{3}3! +\dbinom{4}{4}4! \\ &&&&& = 1!C_4^1 +2!C_4^2 + 3!C_4^3 + 4!C_4^4 \\ &&&&& = 4+12+24+24 \\ &&&&&\mathbf{ = 64} \\ \hline \end{array}\)

-Given that 2a = 3b show that tan theta  = 4/7 and find the corresponding values of theta

\(\begin{array}{|lrcll|} \hline &&& \boxed{a= 2\sin(\theta) + \cos(\theta) \\ b= 2\cos(\theta) - \sin(\theta) } \\\\ &2a &=& 3b \\\\ &2\Big(2\sin(\theta) + \cos(\theta)\Big) &=& 3\Big( 2\cos(\theta) - \sin(\theta)\Big) \\\\ &4\sin(\theta) + 2\cos(\theta) &=& 6\cos(\theta) - 3\sin(\theta) \\\\ &4\sin(\theta) + 3\sin(\theta) &=& 6\cos(\theta) - 2\cos(\theta) \\\\ &7\sin(\theta) &=& 4\cos(\theta) \quad | \quad : \cos(\theta) \\\\ &7\cdot \dfrac{\sin(\theta)}{\cos(\theta)} &=& 4 \quad | \quad : 7 \\\\ &\dfrac{\sin(\theta)}{\cos(\theta)} &=& \dfrac{4}{7} \quad | \quad \dfrac{\sin(\theta)}{\cos(\theta)} = \tan(\theta) \\\\ &\tan(\theta) &=& \dfrac{4}{7} \\\\ & \theta &=& \arctan\left(\dfrac{4}{7}\right) + z\cdot 180^{\circ}, ~ z \in \mathbb{Z} \\\\ & \theta &=& 29.7448812969^{\circ} + z\cdot 180^{\circ} \\\\ z=0: & \mathbf{\theta} &\mathbf{=}& \mathbf{29.7448812969^{\circ}} \\\\ z=1: & \theta &=& 29.7448812969^{\circ}+ 180^{\circ} \\\\ & \mathbf{\theta} &\mathbf{=}& \mathbf{209.744881297^{\circ}} \\ \hline \end{array}\)

The corresponding values of theta are:  \(\mathbf{29.7448812969^{\circ}}\) and \(\mathbf{209.744881297^{\circ}}\)

It is given that a=2sin theta + cos theta b=2cos theta-sin theta,

where theta is greater than or equal to 0 degrees and smaller than or equal to 360 degrees.

-Show that a squared +b squared is constant for all values if theta

\(\begin{array}{|rcll|} \hline && \boxed{a= 2\sin(\theta) + \cos(\theta) \\ b= 2\cos(\theta) - \sin(\theta) } \\\\ \mathbf{a^2+b^2} &=& \Big(2\sin(\theta) + \cos(\theta)\Big)^2 + \Big(2\cos(\theta) - \sin(\theta)\Big)^2 \\\\ &=& 4\sin^2(\theta) + 4\sin(\theta)\cos(\theta) + \cos^2(\theta) \\ &+& 4\cos^2(\theta) - 4\sin(\theta)\cos(\theta) + \sin^2(\theta) \\\\ &=& 4\sin^2(\theta) + \cos^2(\theta)+ 4\cos^2(\theta) + \sin^2(\theta) \\\\ &=& 4\sin^2(\theta)+ 4\cos^2(\theta) + \cos^2(\theta) + \sin^2(\theta) \\\\ &=& 4\Big(\sin^2(\theta)+ \cos^2(\theta)\Big) + \Big(\sin^2(\theta) + \cos^2(\theta)\Big) \\\\ &=& \Big(\sin^2(\theta)+ \cos^2(\theta)\Big)(4+1) \\\\ &=& 5\Big(\sin^2(\theta)+ \cos^2(\theta)\Big) \quad | \quad \sin^2(\theta)+ \cos^2(\theta) = 1 \\\\ &=& 5 \cdot 1 \\\\ &\mathbf{=}& \mathbf{5} \\ \hline \end{array} \)

Prove that  \[1{n\choose 1} + 2{n\choose 2} + 3{n \choose 3} + \cdots +  n{n\choose n} = n2^{n-1}\] 

for all positive integers n.

\(\begin{array}{|rcll|} \hline && \mathbf{1\dbinom n1 + 2\dbinom n2 + 3\dbinom n3 + 4\dbinom n4 + \ldots + n\dbinom nn} \\\\ &=& 1\cdot\dfrac{n!}{(n-1)!\cdot 1} + 2\cdot\dfrac{n!}{(n-2)!\cdot 1\cdot 2} + 3\cdot\dfrac{n!}{(n-3)!\cdot 1\cdot 2\cdot 3} \\ &&+ 4\cdot\dfrac{n!}{(n-4)!\cdot 1\cdot 2\cdot 3\cdot 4} + \ldots + n\cdot\dfrac{n!}{(n-n)!\cdot 1\cdot 2\cdot \ldots \cdot(n-1)\cdot n } \\\\ &=& \dfrac{n!}{(n-1)!\cdot 0!} + \dfrac{n!}{(n-2)!\cdot 1!} + \dfrac{n!}{(n-3)!\cdot 2!} + \dfrac{n!}{(n-4)!\cdot 3!} \\ && +\ldots + \dfrac{n!}{(n-n)!\cdot(n-1)! } \\\\ &&\boxed{n!=n\cdot(n-1)!} \\\\ &=& \dfrac{n\cdot(n-1)!}{(n-1)!\cdot 0!} + \dfrac{n\cdot(n-1)!}{(n-2)!\cdot 1!} + \dfrac{n\cdot(n-1)!}{(n-3)!\cdot 2!} + \dfrac{n\cdot(n-1)!}{(n-4)!\cdot 3!} \\ && +\ldots + \dfrac{n\cdot(n-1)!}{(n-n)!\cdot(n-1)! } \\\\ &=& n\cdot \Big[ \dfrac{ (n-1)!}{(n-1)!\cdot 0!} + \dfrac{ (n-1)!}{(n-2)!\cdot 1!} + \dfrac{ (n-1)!}{(n-3)!\cdot 2!} + \dfrac{ (n-1)!}{(n-4)!\cdot 3!} \\ && +\ldots + \dfrac{ (n-1)!}{(n-n)!\cdot(n-1)! } \Big] \\\\ &=& n\cdot \Big[ \underbrace{\dbinom{n-1}{0}+\dbinom{n-1}{1}+\dbinom{n-1}{2}+\dbinom{n-1}{3} +\ldots + \dbinom{n-1}{n-1}}_{=2^{n-1}} \Big] \\\\ &\mathbf{=}& \mathbf{n\cdot 2^{n-1}} \\ \hline \end{array}\)

If 1,000 mod n =1, where n=1, 2, 3,4...........998, 999, 1000, just how many n are there between 1 and 1,000?

\(\text{Since $1000 \pmod 1 = 0$ and $1000 \pmod {1000} = 0$, $n$ can only be $2,3, \ldots , 999$.} \)

\(\begin{array}{|rcll|} \hline 1000 & \equiv & 1 \pmod n \\ \text{or} \\ 1000-1 &=& n\cdot m,~ \text{with } m \in \mathbb{Z} \\ 999 &=& n\cdot m,~ \text{so $n$ are all divisors of $999$ except $1$ } \\ \hline \end{array} \)

The divisors of 999 are:

Divisors:

1 | 3 | 9 | 27 | 37 | 111 | 333 | 999 (8 divisors)

n = 3 | 9 | 27 | 37 | 111 | 333 | 999 (7 numbers)

Vielen Dank, asinus.

Vollständige Induktion!

Zeigen Sie mit vollständiger Induktion
\(\displaystyle \sum\limits_{k=1}^{n} (-1)^{k-1}\cdot k=\dfrac{1}{4}\left[~1+(-1)^{n-1}\cdot (2n+1)~\right]\)
für alle \(n \in \mathbb{N}\):

\(\bf{\text{Beweise mit vollständiger Induktion:}}\)

\(\displaystyle \sum\limits_{k=1}^{n} (-1)^{k-1}\cdot k=\dfrac{1}{4}\left[~1+(-1)^{n-1}\cdot (2n+1)~\right]\)

\(\text{Induktionsanfang:}\)

\(\begin{array}{|lll|} \hline n=1 & \text{linke Seite:} & (-1)^{1-1}\cdot 1 \\ & &= 1 \\\\ & \text{rechte Seite:} & \dfrac{1}{4}[~1+(-1)^{1-1}\cdot (2\cdot 1+1)~] \\ & &= \dfrac{1}{4}[~1+1\cdot 3~] \\ & &= \dfrac{1}{4}\cdot(4) \\ & &= 1 \\ \hline \end{array} \)

\(\text{Für $\mathbf{n=1}$ sind beide Seiten gleich, und die Aussage ist wahr!}\)

\(\text{Die Induktionsannahme (I.A.) lautet:}\)

\(\begin{array}{|rcll|} \hline \displaystyle \sum\limits_{k=1}^{n} (-1)^{k-1}\cdot k &=& \dfrac{1}{4}\left[~1+(-1)^{n-1}\cdot (2n+1)~\right] \\ \hline \end{array}\)

\(\text{Der Induktionsschluss von $\mathbf{n}$ nach $\mathbf{n+1}$:}\)

\(\begin{array}{|rcll|} \hline \displaystyle \sum\limits_{k=1}^{n+1} (-1)^{k-1}\cdot k &=& \dfrac{1}{4}\left[~1+(-1)^{(n+1)-1}\cdot (2(n+1)+1)~\right] \\ \hline \end{array}\)

\(\bf{\text{linke Seite:}}\)

\(\begin{array}{|llrcll|} \hline &\mathbf{ \sum\limits_{k=1}^{n+1} (-1)^{k-1}\cdot k }\\\\ = & \sum\limits_{k}^{n} (-1)^{k-1}\cdot k + (-1)^{(n+1)-1}(n+1) \\\\ = & \sum\limits_{k}^{n} (-1)^{k-1}\cdot k + (-1)^{n}(n+1) \\\\ \overset{I.A.}{=} & \dfrac{1}{4}\left[~1+(-1)^{n-1}\cdot (2n+1)~\right] + (-1)^{n}(n+1) \\\\ = & \dfrac{1}{4}\left[~1+(-1)^{n-1}\cdot (2n+1)~\right] + (-1)^{n}(n+1) \cdot \dfrac{4}{4} \\\\ = & \dfrac{1}{4}\left[~1+\dfrac{(-1)^{n}}{-1}\cdot (2n+1)+ 4\cdot(-1)^{n}(n+1) ~\right] \\\\ = & \dfrac{1}{4}\left[~1- (-1)^{n}\cdot (2n+1)+ 4\cdot(-1)^{n}(n+1) ~\right] \\\\ = & \dfrac{1}{4}\left\{~1 +(-1)^n\cdot[~ 4(n+1)-(2n+1)~] ~\right\} \\\\ = & \dfrac{1}{4}\left[~1 +(-1)^n\cdot (4n+4-2n-1)~\right] \\\\ \mathbf{=} & \mathbf{\dfrac{1}{4}\left[~1 +(-1)^n\cdot(~ 2n+3) ~\right] }\\ \hline \end{array}\)

\(\bf{\text{rechte Seite:}} \)

\(\begin{array}{|ll|} \hline & \mathbf{\dfrac{1}{4}\left[~1+(-1)^{(n+1)-1}\cdot (2(n+1)+1)~\right] } \\\\ =& \dfrac{1}{4}\left[~1+(-1)^{n}\cdot (2n+2+1)~\right] \\\\ \mathbf{=}& \mathbf{\dfrac{1}{4}\left[~1+(-1)^{n}\cdot (2n+3)~\right]} \\ \hline \end{array}\)