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 #1
avatar+20009 
+3

Vectors:

\(\text{Let $\vec{AN}=2\vec{OA}=2a$} \\ \text{Let $\vec{ON}=\vec{OA}+\vec{AN}=a+2a=3a$} \\ \text{Let $\vec{OM}=\dfrac12\vec{OB}=\dfrac{b}{2}$} \)

 

\(\mathbf{\vec{AB}} =\ ?\)

\(\begin{array}{|rcll|} \hline \vec{OA} + \vec{AB} &=& \vec{OB} \quad & | \quad \vec{OA}=a \qquad \vec{OB}=b \\ a + \vec{AB} &=& b \\ \mathbf{\vec{AB}} &\mathbf{=}& \mathbf{b-a} \\ \hline \end{array}\)

 

\(\mathbf{\vec{MN}} =\ ?\)

\(\begin{array}{|rcll|} \hline \vec{MN} &=& \vec{ON} - \vec{OM} \quad & | \quad \vec{ON}=3a \qquad \vec{OM}=\dfrac{b}{2} \\ \mathbf{\vec{MN}} &\mathbf{=}& \mathbf{3a - \dfrac{b}{2}} \\ \hline \end{array}\)


\(\mathbf{\vec{MP}} =\ ?\)

\(\begin{array}{|rcll|} \hline \vec{OM}+\vec{MP} &=& \vec{OA} + \vec{AP} \quad & | \quad \vec{OA}=a \quad \vec{OM}=\dfrac{b}{2} \quad \vec{AP}=k\vec{AB}\quad \\ \dfrac{b}{2}+\vec{MP} &=& a + k\vec{AB} \quad & | \quad \vec{AB}=b-a \\ \dfrac{b}{2}+\vec{MP} &=& a + k(b-a) \\ \vec{MP} &=& a + k(b-a)-\dfrac{b}{2} \\ \vec{MP} &=& a + kb -ka-\dfrac{b}{2} \\ \vec{MP} &=& a(1-k) + b(k-\dfrac12) \\ \mathbf{\vec{MP}} &\mathbf{=}& \mathbf{a(1-k) + b\left(k-\dfrac12 \right)} \\ \hline \end{array}\)

 

\(\mathbf{k} =\ ? \\ \text{Let $\vec{MP}=\lambda\vec{MN}$, where $\lambda$ is a scalar quantity.} \\ \text{Let $\vec{MN}=\vec{ON}-\vec{OM}=3a-\dfrac{b}{2}$} \)

\(\begin{array}{|rcll|} \hline \vec{OM}+\vec{MP} &=& \vec{OA} + \vec{AP} \quad & | \quad \vec{OA}=a \quad \vec{OM}=\dfrac{b}{2} \quad \vec{AP}=k\vec{AB}\quad \\ \dfrac{b}{2}+\vec{MP} &=& a + k\vec{AB} \quad & | \quad \vec{MP}= \lambda\vec{MN} \quad \vec{AB}=b-a \\ \dfrac{b}{2}+\lambda\vec{MN} &=& a + k(b-a) \quad & | \quad \vec{MN}= 3a-\dfrac{b}{2} \\ \dfrac{b}{2}+\lambda \left(3a-\dfrac{b}{2} \right) &=& a + k(b-a) \\ \dfrac{b}{2}+ 3\lambda a-\dfrac{\lambda}{2}b &=& a + kb-ka \\ 3\lambda a -a+ka &=& kb + \dfrac{\lambda}{2}b -\dfrac{b}{2} \\ a\underbrace{\left( 3\lambda -1+k \right)}_{=0} &=& b\underbrace{\left( k + \dfrac{\lambda}{2} -\dfrac{1}{2} \right)}_{=0} \quad & | \quad \text{linearly independent vectors }~a \text{ and } ~ b \\ \hline k + \dfrac{\lambda}{2} -\dfrac{1}{2} &=& 0 \quad & | \quad \cdot 2 \\ 2k + \lambda -1 &=& 0 \\ \mathbf{ \lambda } & \mathbf{=} & \mathbf{1-2k} \\ \hline 3\lambda -1+k &=& 0 \quad & | \quad \lambda=1-2k \\ 3(1-2k) -1+k &=& 0 \\ 3-6k -1+k &=& 0 \\ 2-5k &=& 0 \\ 5k &=& 2\\ \mathbf{k} &\mathbf{=}& \mathbf{\dfrac25} \\ \hline \end{array}\)

 

 

 

laugh

heureka 19.09.2018
 #1
avatar+20009 
+2

A closed form for the sum $$ S = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1} $$
is $1 - \frac{a^{n+b}}{3^{2^{n+c}}-1},$ where $a$, $b$, and $c$ are integers. Find $a+b+c$.

 

Formula:

\(\begin{array}{|rclcl|} \hline s_n &=& \dfrac{2}{3+1} + \dfrac{2^2}{3^2+1} + \dfrac{2^3}{3^{(2^2)}+1} + \dfrac{2^4}{3^{(2^3)}+1} + \dfrac{2^5}{3^{(2^4)}+1} + \ldots + \dfrac{2^{n+1}}{3^{(2^n)}+1} \\\\ s_0 &=& \dfrac12 \\ s_1 &=& \dfrac12 + \dfrac{4}{10} = \dfrac{9}{10} \\ s_2 &=& \dfrac{9}{10}+ \dfrac{8}{82}= \dfrac{409}{410} \\ s_3 &=& \dfrac{409}{410}+ \dfrac{16}{6562}= \dfrac{1345209}{1345210} \\ \ldots \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline s_n &=& 1 - \dfrac{a^{n+b}}{3^{ (2^{n+c} )}-1} \\\\ s_n &=& 1 - \dfrac{a^n a^b} { 3^{ (2^n 2^c )}-1 } \\\\ s_n &=& 1 - \dfrac{a^n a^b} { \left(3^{(2^c)} \right)^{2^n}-1 } \\ && \text{Set }~ \alpha=a^b \\ && \text{Set }~ \gamma=3^{(2^c)} \\ \mathbf{s_n} & \mathbf{=} & \mathbf{1 - \dfrac{a^n \alpha} { \gamma^{(2^n)}-1 } } \\ \hline \end{array}\)

 

\(\begin{array}{rcll} \boxed{n=0}:\\ s_0 = \dfrac12 &=& 1 - \dfrac{a^0 \alpha} { \gamma^{(2^0)}-1 } \\ \dfrac12 &=& 1 - \dfrac{\alpha} { \gamma-1 } \\ \dfrac{\alpha} { \gamma-1 } &=& 1 - \dfrac12 \\ \dfrac{\alpha} { \gamma-1 } &=& \dfrac12 \\ \mathbf{ \alpha } & \mathbf{=}& \mathbf{ \dfrac12\cdot (\gamma-1)} \qquad (1) \\ \end{array} \)

 

\(\begin{array}{rcll} \boxed{n=1}:\\ s_1 = \dfrac{9}{10} &=& 1 - \dfrac{a^1 \alpha} { \gamma^{(2^1)}-1 } \\ \dfrac{9}{10} &=& 1 - \dfrac{a \alpha} { \gamma^2-1 } \\ \dfrac{a \alpha} { \gamma^2-1 } &=& 1 - \dfrac{9}{10} \\ a\alpha & = & \dfrac{1}{10}\cdot (\gamma^2-1) \quad & | \quad \gamma^2-1 = (\gamma-1)(\gamma+1) \\ \mathbf{ a\alpha } & \mathbf{=}& \mathbf{ \dfrac{1}{10}\cdot (\gamma-1)(\gamma+1)} \qquad (2) \\ \end{array}\)

 

\(\begin{array}{rcll} \boxed{n=2}:\\ s_2 = \dfrac{409}{410} &=& 1 - \dfrac{a^2 \alpha} { \gamma^{(2^2)}-1 } \\ \dfrac{409}{410} &=& 1 - \dfrac{a^2 \alpha} { \gamma^4-1 } \\ \dfrac{a^2 \alpha} { \gamma^4-1 } &=& 1 - \dfrac{409}{410} \\ a^2 \alpha &=& \dfrac{1}{410}(\gamma^4-1) \quad | \quad \gamma^4-1 = (\gamma^2-1)(\gamma^2+1)= (\gamma-1)(\gamma+1)(\gamma^2+1) \\ \mathbf{ a^2\alpha } & \mathbf{=}& \mathbf{ \dfrac{1}{410}\cdot (\gamma-1)(\gamma+1)(\gamma^2+1)} \qquad (3) \\ \end{array}\)

 

\(\mathbf{a=~ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{ a\alpha } & \mathbf{=}& \mathbf{ \dfrac{1}{10}\cdot (\gamma-1)(\gamma+1)} \qquad (2) \quad & | \quad \mathbf{ \alpha = \dfrac12\cdot (\gamma-1)} \qquad (1) \\ a\cdot \dfrac12\cdot (\gamma-1) & = & \dfrac{1}{10}\cdot (\gamma-1)(\gamma+1) \\ a\cdot \dfrac12 & = & \dfrac{1}{10}\cdot (\gamma+1) \\ \mathbf{a} & \mathbf{=} & \mathbf{\dfrac{1}{5}\cdot (\gamma+1)} \qquad (4) \\ \hline \end{array} \)


\(\mathbf{\gamma=~ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{ a^2\alpha } & \mathbf{=}& \mathbf{ \dfrac{1}{410}\cdot (\gamma-1)(\gamma+1)(\gamma^2+1)} \qquad (3) \\\\ && \mathbf{a = \dfrac{1}{5}\cdot (\gamma+1)} \quad (4) \qquad \mathbf{ \alpha = \dfrac12\cdot (\gamma-1)} \quad (1)\\\\ \dfrac{1}{25}\cdot (\gamma+1)^2\cdot \dfrac12\cdot (\gamma-1) & = & \dfrac{1}{410}\cdot (\gamma-1)(\gamma+1)(\gamma^2+1) \\ \dfrac{1}{25}\cdot (\gamma+1)\cdot \dfrac12 & = & \dfrac{1}{410}\cdot (\gamma^2+1) \\ \gamma+1 & = & \dfrac{5}{41}\cdot (\gamma^2+1) \\ \gamma+1 & = & \dfrac{5}{41}\gamma^2+\dfrac{5}{41} \\ \dfrac{5}{41}\gamma^2-\gamma-1+\dfrac{5}{41} &=& 0 \\ \dfrac{5}{41}\gamma^2-\gamma- \dfrac{36}{41} &=& 0 \\\\ \gamma &=& \dfrac{ 1\pm \sqrt{1-4\cdot\dfrac{5}{41}\cdot \left(-\dfrac{36}{41} \right) } } {2\cdot \dfrac{5}{41} } \\\\ \gamma &=& \dfrac{ 1\pm \sqrt{1+ \dfrac{720}{41^2} } } {\dfrac{10}{41} } \\\\ \gamma &=& \dfrac{41}{10} \left( 1\pm \sqrt{1+ \dfrac{720}{41^2} } \right) \\\\ \gamma &=& \dfrac{41}{10} \left( 1\pm \dfrac{ \sqrt{2401} } {41} \right) \\\\ \gamma &=& \dfrac{41}{10} \left( 1\pm \dfrac{ 49 } {41} \right) \\\\ \gamma &=& \dfrac{41}{10} \pm \dfrac{41}{10} \cdot \dfrac{49} {41} \\\\ \gamma &=& \dfrac{41}{10} \pm \dfrac{49}{10} \\\\ \gamma &=& \dfrac{41+49}{10} \\\\ \gamma &=& \dfrac{90}{10} \\ \mathbf{ \gamma }& \mathbf{=}& \mathbf{9} \\\\ \gamma &=& \dfrac{41-49}{10} \\\\ \gamma &=& -\dfrac{8}{10} \qquad \text{no solution, } \gamma\gt 0\\ \hline \end{array} \)

 

\(\mathbf{a=~ ?}\)

\(\begin{array}{|rcll|} \hline a & = & \dfrac15\cdot (\gamma+1) \quad & | \quad \gamma = 9 \\ a & = & \dfrac15\cdot (9+1) \\ a & = & \dfrac{10}{5} \\ \mathbf{a} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)

 

\(\mathbf{\alpha=~ ?}\)

\(\begin{array}{|rcll|} \hline \alpha & = & \dfrac12\cdot (\gamma-1) \quad & | \quad \gamma = 9 \\ \alpha & = & \dfrac12\cdot (9-1) \\ \alpha & = & \dfrac{8}{2} \\ \mathbf{\alpha} & \mathbf{=} & \mathbf{4} \\ \hline \end{array}\)

 

\(\mathbf{b=~ ?}\)

\(\begin{array}{|rcll|} \hline \alpha &=& a^b \quad & | \quad a=2 \qquad \alpha = 4 \\ 4 &=& 2^b \\ 2^2 &=& 2^b \\ 2 &=& b \\ \mathbf{b} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)

 

\(\mathbf{c=~ ?} \)

\(\begin{array}{|rcll|} \hline \gamma &=& 3^{(2^c)} \quad & | \quad \gamma = 9 \\ 9 &=& 3^{(2^c)} \\ 3^2 &=& 3^{(2^c)} \\ 2 &=& 2^c \\ 2^1 &=& 2^c \\ 1 &=& c \\ \mathbf{c} & \mathbf{=} & \mathbf{1} \\ \hline \end{array}\)

 

\(\begin{array}{rcll} && \mathbf{a+b+c} \\ &=& 2+2+1 \\ &=& \mathbf{5} \\ \end{array}\)

 

laugh

heureka 17.09.2018
 #3
avatar+20009 
+1

b)

(-2x) * (4-3x^2)^-2 dx Grenzen 1 und -1 durch Substitution.

 

Integral

\(\displaystyle \int \limits_{-1}^{1} \dfrac{-2x}{ (4-3x^2)^2 } \ dx = \ ?\)

 

\(\begin{array}{|lrcl|} \hline \displaystyle \int \limits_{-1}^{1} \dfrac{-2x}{ (4-3x^2)^2 } \ dx = \ ? \\ & \text{Substitution:}\\ & \mathbf{z} &\mathbf{=}& \mathbf{ 4-3x^2 } \\ & dz &=& -6x \ dx \\ & \mathbf{dx} &\mathbf{=}& \mathbf{-\dfrac{1} { 6x } \ dz} \\ \hline \end{array}\)

 

Stammfunktion:

\(\begin{array}{|lcl|} \hline \displaystyle \int \dfrac{-2x}{ (4-3x^2)^2 } \ dx \\\\ &=& \displaystyle \int \left( \dfrac{-2x}{ z^2 }\right) \cdot \left(-\dfrac{1} { 6x } \right) \ dz \\\\ &=& \displaystyle \dfrac13\cdot \int \dfrac{1}{ z^2 } \ dz \\\\ &=& \displaystyle \dfrac13\cdot \int z^{-2} \ dz \\\\ &=& \displaystyle \dfrac13\cdot\left( \dfrac{z^{-2+1} }{-2+1} \right) +c \\\\ &=& \displaystyle -\dfrac13\cdot z^{-1} +c \\\\ &=& \displaystyle -\dfrac{1}{3z} +c \\\\ & \text{Rücksubstitution:} \\ &=& \displaystyle -\dfrac{1}{3(4-3x^2)} +c \\\\ \hline \mathbf{\displaystyle \int \limits_{-1}^{1} \dfrac{-2x}{ (4-3x^2)^2 } \ dx} \\ &=& \displaystyle -\dfrac13 \cdot \left[ \dfrac{1}{4-3x^2} \right]_{-1}^{1} \\\\ &=& \displaystyle -\dfrac13 \cdot \left[ \dfrac{1}{4-3\cdot 1^2} - \dfrac{1}{4-3\cdot (-1)^2} \right] \\\\ &=& \displaystyle -\dfrac13 \cdot \left[ \dfrac{1}{1} - \dfrac{1}{1} \right] \\\\ &=& \displaystyle -\dfrac13 \cdot \left[ 1 - 1 \right] \\\\ &=& \displaystyle -\dfrac13 \cdot 0 \\\\ &=& \displaystyle 0 \\\\ \mathbf{\displaystyle \int \limits_{-1}^{1} \dfrac{-2x}{ (4-3x^2)^2 } \ dx} &\mathbf{=}& \mathbf{0} \\ \hline \end{array}\)

 

 

laugh

heureka 13.09.2018
 #2
avatar+20009 
+1
heureka 13.09.2018