heureka

What are the first 4 terms of f(n-1)+15

If f(1) = 7

$\begin{array}{rcll} f(2) &=& f(2-1)+15 \\ f(2) &=& f(1)+15 \qquad | \qquad f(1) = 7\\ f(2) &=& 7+15\\ f(2) &=& 22 \\\\ f(3) &=& f(3-1)+15 \\ f(3) &=& f(2)+15 \qquad | \qquad f(2) = 22\\ f(3) &=& 22+15\\ f(3) &=& 37 \\\\ f(4) &=& f(4-1)+15 \\ f(4) &=& f(3)+15 \qquad | \qquad f(3) = 37\\ f(4) &=& 37+15\\ f(4) &=& 52 \\\\ \cdots \end{array}$

arithmetic sequence: $\boxed{ ~ a_n=a_1+(n-1)\cdot d ~}$

$\begin{array}{rcll} f(n) &=& f(n-1)+15 \\ f(n)- f(n-1) &=& 15 \\ d &=& 15\\\\ \end{array}$

$\begin{array}{rcll} a_1 = f(1) &=& 7 \\ d &=& 15 \\\\ a_n &=& 7 + (n-1)\cdot 15 \\ a_n &=& 7 + 15n -15 \\ \mathbf{ a_n} & \mathbf{ = } & \mathbf{ -8 + 15n } \end{array}$

$\begin{array}{rcll} \int \limits_{1}^{3} \frac{x-4}{2x+1} \ dx = \ ? \end{array}$

$\begin{array}{rcll} \int \limits_{1}^{3} \frac{x-4}{2x+1} \ dx &= & \int \limits_{1}^{3} \frac{x}{2x+1} \ dx -4 \int \limits_{1}^{3} \frac{1}{2x+1} \ dx \\ &= & \int \limits_{1}^{3} \frac{x}{2(x+\frac12)} \ dx -4 \int \limits_{1}^{3} \frac{1}{2x+1} \ dx \\ &= & \frac12 \int \limits_{1}^{3} \frac{x}{x+\frac12} \ dx -4 \int \limits_{1}^{3} \frac{1}{2x+1} \ dx \\ &= & \frac12 \int \limits_{1}^{3} \frac{x+\frac12-\frac12}{x+\frac12} \ dx -4 \int \limits_{1}^{3} \frac{1}{2x+1} \ dx \\ &= & \frac12 \int \limits_{1}^{3} \ dx +\frac12 \int \limits_{1}^{3} \frac{-\frac12}{x+\frac12} \ dx -4 \int \limits_{1}^{3} \frac{1}{2x+1} \ dx \\ &= & \frac12 \int \limits_{1}^{3} \ dx -\frac14 \int \limits_{1}^{3} \frac{1}{x+\frac12} \ dx -4 \int \limits_{1}^{3} \frac{1}{2x+1} \ dx \\ &= & \frac12 \int \limits_{1}^{3} \ dx -\frac94 \int \limits_{1}^{3} \frac{1}{x+\frac12} \ dx \\ &= & \frac12 [x]_1^3 -\frac94 [\ln{(x+\frac12)}]_1^3 \\ &= & \frac12 (3-1) -\frac94 [\ln{(3+\frac12)}-\ln{(1+\frac12)} ] \\ &= & \frac12 (3-1) -\frac94 [\ln{ (\frac72)}-\ln{(\frac32)} ] \\ &= & \frac12 (3-1) -\frac94 [\ln{ \left( \frac{ \frac72 }{ \frac32 } \right) } ] \\ &= & \frac12 (3-1) -\frac94 [\ln{ ( \frac{ 7 }{ 3 } ) } ] \\ &= & \frac12 (2) -\frac94 [\ln{ ( \frac{ 7 }{ 3 } ) } ] \\ &= & 1 -\frac94 [\ln{ ( \frac{ 7 }{ 3 } ) } ] \\ &=& 1-2,25\cdot 0,84729786039\\ &=& 1- 1,90642018587\\ &=& -0,90642018587\\ \mathbf{ \int \limits_{1}^{3} \frac{x-4}{2x+1} \ dx } &\mathbf{=}& \mathbf{-0,90642018587} \end{array}$

Alpha + beta = 270 , cos alpha + sin beta = 0

$\small{ \begin{array}{rcll} \alpha + \beta &=& 270 \\ \beta &=& 270 - \alpha \\\\ \cos{(\alpha)} + \sin{(\beta)} &=& \cos{(\alpha)} +\sin{(270 - \alpha)} \\ && \sin{(270 - \alpha)} = \sin{(270 )}\cos{(\alpha)} + \cos{(270 )}\sin{(\alpha)} \\ &=& \cos{(\alpha)} +\sin{(270 )}\cos{(\alpha)} + \cos{(270 )}\sin{(\alpha)} \quad & | \quad \sin{(270 )} = -1 \qquad \cos{(270 )} = 0 \\ &=& \cos{(\alpha)} +(-1)\cdot \cos{(\alpha)} + 0\cdot \sin{(\alpha)}\\ &=& \cos{(\alpha)} -\cdot \cos{(\alpha)} \\ &=& 0 \\ \cos{(\alpha)} + \sin{(\beta)} &=& 0 \end{array} }$

write the next term in this sequence

-1 2 7 14 23...

$\begin{array}{lrrrrrrrrrr} & {\color{red}d_0 = -1} && 2 && 7 && 14 && 23 && \cdots \\ \text{1. Difference } && {\color{red}d_1 = 3} && 5 && 7 && 9 && 11 && \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = 2} && 2 && 2 && 2 && 2 && \cdots \\ \end{array}$

$\begin{array}{rcl} a_n &=& \binom{n-1}{0}\cdot {\color{red}d_0 } + \binom{n-1}{1}\cdot {\color{red}d_1 } + \binom{n-1}{2}\cdot {\color{red}d_2 }\\ a_n &=& \binom{n-1}{0}\cdot ({\color{red}-1 }) + \binom{n-1}{1}\cdot {\color{red}3} + \binom{n-1}{2}\cdot {\color{red}2}\\ \\ \hline \binom{n-1}{0} &=& 1 \\ \binom{n-1}{1} &=& n-1 \\ \binom{n-1}{2} &=& ( \frac{n-1}{2} ) \cdot ( \frac{n-2}{1} ) \\ \hline \\ a_n &=& ({\color{red}-1 }) + (n-1)\cdot {\color{red}3} + ( \frac{n-1}{2} ) \cdot ( \frac{n-2}{1} )\cdot {\color{red}2}\\ a_n &=& -1 + 3n - 3 + (n-1)(n-2) \\ a_n &=& -1 + 3n - 3 + n^2-2n-n+2 \\ \mathbf{a_n} &=& \mathbf{n^2-2} \\\\ a_6 &=& 6^2-2 = 34\\ a_7 &=& 7^2-2 = 47\\ a_8 &=& 8^2-2 = 62\\ \cdots \end{array}$

On a certain college faculity, 4/7 of the professors are male, and the ratio of the professors older than 50 to the professors less than or equal to 50 years is 2:5. if 1/5 of the male professors are older than 50 years, what fraction of the female professors are less than or equal to 50 years ?

$\begin{array}{rcl} m_{>50} &=& \frac47\cdot \frac15 = \frac{4}{35} \\ m_{\le50} &=& \frac47\cdot \frac45 = \frac{16}{35} \\ m_{>50} + m_{\le50} &=& \frac{4}{35} + \frac{16}{35} = \frac{20}{35} \\\\ f_{>50} + f_{\le50} &=& 1- ( m_{>50} + m_{\le50} ) = 1- \frac{20}{35} = \frac{15}{35} \\ f_{>50} &=& \frac{15}{35} - f_{\le50} \\\\ \frac{m_{>50}+f_{>50}}{m_{\le50}+f_{\le50}} &=& \frac25 \\ 5\cdot ( m_{>50}+f_{>50} ) &=& 2\cdot ( m_{\le50}+f_{\le50} ) \qquad | \qquad f_{>50} = \frac{15}{35} - f_{\le50} \\ 5\cdot ( m_{>50}+\frac{15}{35} - f_{\le50} ) &=& 2\cdot ( m_{\le50}+f_{\le50} ) \\ 5\cdot ( \frac{4}{35}+\frac{15}{35} - f_{\le50} ) &=& 2\cdot ( \frac{16}{35}+f_{\le50} ) \\ 5\cdot ( \frac{19}{35} - f_{\le50} ) &=& 2\cdot ( \frac{16}{35}+f_{\le50} ) \\ \frac{95}{35} - 5\cdot f_{\le50} &=& \frac{32}{35}+2\cdot f_{\le50} \\ 7\cdot f_{\le50} &=& \frac{95}{35} - \frac{32}{35} \\ 7\cdot f_{\le50} &=& \frac{63}{35} \\ f_{\le50} &=& \frac{63}{35\cdot 7 } \\ \text{female}_{\le50} &=& \frac{9}{35} \\ \end{array}$

what is 111...1(total of one hundred ones) divided by 1,111,111 ( answer with remainder, not fraction or decimal )

You can divide the number in  arbitrarily sections.

Devide all parts by 1 111 111, but attach the remainder of the previous calculation left.

Example:

$\underbrace{1111111}_{1.\ \text{partition}}\ \underbrace{1111111}_{2.\ \text{partition}}\ \underbrace{1111111}_{3.\ \text{partition}}\ \underbrace{1111111}_{4.\ \text{partition}}\ \underbrace{1111111}_{5.\ \text{partition}}\ \underbrace{1111111}_{6.\ \text{partition}}\ \underbrace{1111111}_{7.\ \text{partition}}\ \\ \underbrace{1111111}_{8.\ \text{partition}}\ \underbrace{1111111}_{9.\ \text{partition}}\ \underbrace{1111111}_{10.\ \text{partition}}\ \underbrace{1111111}_{11.\ \text{partition}}\ \underbrace{1111111}_{12.\ \text{partition}}\ \underbrace{1111111}_{13.\ \text{partition}}\ \underbrace{1111111}_{14.\ \text{partition}}\ \underbrace{11}_{15.\ \text{partition}}\ $

$\begin{array}{lcl} \underbrace{1111111}_{1.\ \text{partition}} : 1\ 111\ 111 = 1 \quad \text{ Remainder } = \color{red}{0} \\ \text{attach Remainder}\\ {\color{red}0}\underbrace{1111111}_{2.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = \color{green}{0} \\ \text{attach Remainder}\\ {\color{green}0}\underbrace{1111111}_{3.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{4.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{5.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{6.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{7.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{8.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{9.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{10.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{11.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{12.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{13.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{1111111}_{14.\ \text{partition}}: 1\ 111\ 111 = 1 \quad \text{ Remainder } = 0 \\ \text{attach Remainder}\\ 0\underbrace{11}_{15.\ \text{partition}}: 1\ 111\ 111 = 0 \quad \text{ Remainder } = \mathbf{11} \\ \end{array}$

Berechnung der Bogenlänge einer Kurve, es gilt

$L(c) = \int \limits_{a}^{b} ||~ \dot c(t) ~|| \ dt$

$\begin{array}{lcl} \text{Für die Polarkoordinaten } r \equiv r(t) \text{ und } \varphi \equiv \varphi (t) \text{ gilt: }\\ \quad c(t) = \binom{\cos{(t)}}{\sin{(t)}} = ( \cos{(t)},\sin{(t)})^T \qquad \text{ für } a\le t \le b \\ \quad L(c) = \int \limits_{a}^{b} \sqrt{ \dot r^2 + r^2\cdot \dot\varphi^2 } \ dt. \end{array}$

Wir haben eine Kardioide (Herzlinie)  in Polarkoordinaten

$\begin{array}{rcl} L(c) &=& \int \limits_{0}^{2\pi} ||~ [~a\cdot(1+\cos{(t)}) ~]' ~|| \ dt \\\\ r(t) &=& a\cdot(1+\cos{(t)}) \\ \varphi(t) &=& t \\\\ \dot r(t) &=& -a\cdot \sin{(t)}\\ \left[ \dot r(t) \right]^2 &=& \dot r^2 = a^2\cdot \sin^2{(t)}\\ \left[ r(t)\right]^2 &=& r^2= a^2\cdot(1+\cos{(t)})^2 \\ \dot \varphi(t) &=& \dot\varphi = 1 \\ \left[ \dot \varphi(t) \right]^2 &=& \dot\varphi^2 = 1^2 = 1 \\ a&=& 0\\ b&=& 2\pi\\\\ L(c) &=& \int \limits_{a}^{b} \sqrt{ \dot r^2 + r^2\cdot \dot\varphi^2 } \ dt\\ L(c) &=& \int \limits_{0}^{2\pi} \sqrt{ \dot r^2 + r^2\cdot \dot\varphi^2 } \ dt\\ L(c) &=& \int \limits_{0}^{2\pi} \sqrt{ a^2\cdot \sin^2{(t)} + a^2\cdot(1+\cos{(t)})^2\cdot 1 } \ dt\\ &=& \int \limits_{0}^{2\pi} \sqrt{ a^2\cdot \sin^2{(t)} + a^2\cdot(1+2\cos{(t)}+\cos^2{(t)} )} \ dt\\ &=& a\cdot \int \limits_{0}^{2\pi} \sqrt{ \sin^2{(t)} + 1+2\cos{(t)}+\cos^2{(t)} } \ dt\\ &=& a\cdot \int \limits_{0}^{2\pi} \sqrt{ \sin^2{(t)} + \cos^2{(t)} + 1+2\cos{(t)} } \ dt\\ &=& a\cdot \int \limits_{0}^{2\pi} \sqrt{ 1 + 1+2\cos{(t)} } \ dt\\ &=& a\cdot \int \limits_{0}^{2\pi} \sqrt{2(1+\cos{(t)}) } \ dt\\ &=& a\cdot \sqrt{2} \int \limits_{0}^{2\pi} \sqrt{ 1+\cos{(t)} } \ dt\\\\ && \boxed{~ \begin{array}{rcl} \text{Formel: } \\ \cos{(2\alpha)} &=& 2\cos^2{(\alpha)}-1\\ \cos{(\alpha)} &=& 2\cos^2{(\frac{\alpha}{2})}-1\\ 1+\cos{(\alpha)} &=& 2\cos^2{(\frac{\alpha}{2})}\\ \sqrt{1+\cos{(\alpha)}} &=& \sqrt{2}\cdot \cos{(\frac{\alpha}{2})}\\ \end{array} ~}\\\\ &=& a\cdot \sqrt{2} \int \limits_{0}^{2\pi} \sqrt{2}\cdot |~\cos{(\frac{t}{2})}~| \ dt\\ &=& 2a\cdot \int \limits_{0}^{2\pi} |~\cos{(\frac{t}{2})}~| \ dt\\ &=& 2a\cdot\left[ 2\cdot \int \limits_{0}^{\pi} \cos{(\frac{t}{2})}\ dt \right]\\ &=& 4a\cdot \int \limits_{0}^{\pi} \cos{(\frac{t}{2})}\ dt \\ &=& 4a\cdot \left[ 2\cdot \sin{(\frac{t}{2})} \right]_{0}^{\pi}\\ &=& 8a\cdot \left[ \sin{(\frac{\pi}{2})} - \sin{(\frac{0}{2})} \right]\\ &=& 8a\cdot \left[ 1 - 0 \right]\\ L(c) &=& 8a\\ \end{array}$

Die  Kardioide (Herzlinie) hat die Bogenlänge von 8a

Wie kann man √4:1 in Geometrie mit einer Zeichnung darstellen?

Hallo aditya@calc.com,

sorry i have no closed solution. Here my way so far.

$\small{ \begin{array}{rcl} 3^{2n+2} -8n -9 &=& 3^{2n}\cdot 3^2-8n-9 \\ &=& 3^{2n}\cdot 9-8n-9 \\ &=& (3^2)^n\cdot 9-8n-9 \\ &=& 9^n\cdot 9-8n-9 \\ &=& 9^{n+1}-8n-9 \\ &=& (1+8)^{n+1}-8n-9 \\ &=& -8n-9 + (1+8)^{n+1}\\ &=& -8n-9 + \underbrace{\binom{n+1}{0}+ \binom{n+1}{1}\cdot 8}_{=8n+9} + \binom{n+1}{2}\cdot 8^2 + \binom{n+1}{3}\cdot 8^3 + \binom{n+1}{4}\cdot 8^4 +\cdots \\ && + \binom{n+1}{n-1}\cdot 8^{n-1} + \binom{n+1}{n}\cdot 8^{n} + \binom{n+1}{n+1}\cdot 8^{n+1} \\\\ &=& \binom{n+1}{2}\cdot 8^2 + \binom{n+1}{3}\cdot 8^3 + \binom{n+1}{4}\cdot 8^4 +\cdots + \binom{n+1}{n-1}\cdot 8^{n-1} + \binom{n+1}{n}\cdot 8^{n} + \binom{n+1}{n+1}\cdot 8^{n+1} \\\\ &=& \binom{n+1}{2}\cdot 2^6 + \binom{n+1}{3}\cdot 2^9 + \binom{n+1}{4}\cdot 2^{12} +\cdots + \binom{n+1}{n-1}\cdot 2^{3n-3} + \binom{n+1}{n}\cdot 2^{3n} + \binom{n+1}{n+1}\cdot 2^{3n+3} \\ \end{array} }$

It is not easy to factorize the binoms $\binom{n+1}{2},~\binom{n+1}{3},~\binom{n+1}{3}\cdots$  to get the exponent of the prim number 2 for every n

but m mimimum is 6, because we have $2^6$

It seems that (no proof !!!!)

  $\begin{array}{rcl} \binom{n+1}{2}\cdot 2^6 &=& 2^m\cdot p \\\\ (n+1)\cdot n \cdot 2^5 &=& 2^m\cdot p \\ \end{array}$

Example:

$\begin{array}{|r|c|l|} \hline n & m & p \\ \hline 1 & 6 & 1 \\ 2 & 6 & 3 \\ \hline 3 & 7 & 3 \\ 4 & 7 & 5 \\ \hline 5 & 6 & 15 \\ 6 & 6 & 21 \\ \hline 7 & 8 & 7 \\ 8 & 8 & 9 \\ \hline 9 & 6 & 45 \\ 10 & 6 & 55 \\ \hline 11 & 7 & 33 \\ 12 & 7 & 39 \\ \hline 13 & 6 & 91 \\ 14 & 6 & 105 \\ \hline 15 & 9 & 15 \\ 16 & 9 & 17 \\ \hline 17 & 6 & 153 \\ 18 & 6 & 171 \\ \hline \cdots \\ \hline \end{array}$

Write the slope-intercept form of the equation of the line described.

A line through (-4,0) Perpendicular to y=x-1

The slope-intercept form of a line is y = mx + b, where m represents the slope and b represents the y-intercept.

The equation y=x-1 shows that the slope of the line m is 1.

The line perpendicular has the slope $m_{\text{perpendicular}} = -\frac{1} {m} = -\frac{1}{1} = -1$

The line through (-4,0) is  $0 = (-1)\cdot (-4)+b$

so  $b=-4$

The slope-intercept form of the equation of the line is  $\mathbf{y = -x - 4}$