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Questions 11
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 #4
avatar+20025 
+1
heureka Sep 26, 2018
 #2
avatar+20025 
+2

Suppose that \(x,y,z\) are positive integers satisfying \(x \le y \le z\) ,
and such that the product of all three numbers is twice their sum.
What is the sum of all possible values of \(z\)?

 

\(\mathbf{z=\ ?}\)

\(\begin{array}{|rcll|} \hline 2(x+y+z) &=& xyz \\ 2(x+y)+2z &=& xyz \\ xyz-2z &=& 2(x+y) \\ z(xy-2) &=& 2(x+y) \\ \mathbf{z} &\mathbf{=} & \mathbf{ \dfrac{2(x+y)}{xy-2} }\\ \hline \end{array}\)

 

\(\mathbf{x \le y \le z}\)

\(\begin{array}{|ll|} \hline x \le y \le \dfrac{2(x+y)}{xy-2} \quad & | \quad x,y,z \gt 0 \\ & \small{\text{The denominator must be greather than zero:}} \\ & \begin{array}{|rcl|} \hline xy-2 &>& 0 \\ xy &>& 2 \\ \mathbf{y} & \mathbf{>} & \mathbf{\dfrac{2}{x}} \quad\text{ or } \quad \mathbf{\dfrac{2}{x} < y }\\ \hline \end{array}\\ \dfrac{2}{x} \lt y \le \dfrac{2(x+y)}{xy-2} \\\\ & \begin{array}{|rcll|} \hline y &\le& \dfrac{2(x+y)}{xy-2} \quad & | \quad \cdot (xy-2) \\ y(xy-2) &\le& 2(x+y) \\ xy^2 -2y &\le& 2x+2y \\ xy^2 -4y -2x &\le& 0 \\ xy^2 -4y -2x &=& 0 \\\\ y &=& \dfrac{4\pm \sqrt{16-4x(-2x)} }{2x} \\ y &=& \dfrac{4\pm \sqrt{16+8x^2} }{2x} \\ y &=& \dfrac{4\pm \sqrt{4(4+2x^2)} }{2x} \\ y &=& \dfrac{4\pm 2\sqrt{ 4+2x^2 } }{2x} \\ y &=& \dfrac{4}{2x} + \dfrac{2}{2x} \sqrt{ 4+2x^2 } \quad & | \quad y > 0 ! \\ \mathbf{ y } & \mathbf{=} & \mathbf{\dfrac{2}{x} + \dfrac{\sqrt{ 4+2x^2 }}{x} } \\ \hline \end{array}\\ \dfrac{2}{x} \lt y \le \dfrac{2}{x} + \dfrac{\sqrt{ 4+2x^2 }}{x} \\ \hline \end{array} \)

 

\(\mathbf{x \text{ and } y =\ ?}\)

\(\begin{array}{|lrcll|} \hline \mathbf{x=1}: & \dfrac{2}{1} \lt &y& \le \dfrac{2}{1} + \dfrac{\sqrt{ 4+2\cdot 1^2 }}{1} \\\\ & 2 \lt &y& \le 2 + \sqrt{ 6 } \\ & 2 \lt &y& \le 2 + 2.4\ldots \\ & 2 \lt &y& \le 4 \\ & && \Rightarrow 2 \lt 3 \le 4 & \mathbf{y = 3} \\ & && \Rightarrow 2 \lt 4 \le 4 & \mathbf{y = 4} \\\\ \mathbf{x=2}: & \dfrac{2}{2} \lt &y& \le \dfrac{2}{2} + \dfrac{\sqrt{ 4+2\cdot 2^2 }}{2} \\\\ & 1 \lt &y& \le 1 + \dfrac{ \sqrt{ 12 } }{2} \\ & 1 \lt &y& \le 1 + 1.7\ldots \\ & 1 \lt &y& \le 2 \\ & && \Rightarrow 1 \lt 2 \le 2 & \mathbf{y = 2} \\\\ \mathbf{x=3}: & \dfrac{2}{3} \lt &y& \le \dfrac{2}{3} + \dfrac{\sqrt{ 4+2\cdot 3^2 }}{3} \\\\ & \dfrac{2}{3} \lt &y& \le \dfrac{2}{3} + \dfrac{ \sqrt{ 22 }}{3} \\ & \dfrac{2}{3} \lt &y& \le \dfrac{2}{3} + 1.5\ldots \\ & \dfrac{2}{3} \lt &y& \le 2.23013858661 \\ & \dfrac{2}{3} \lt &y& \le 2 \\ & && \Rightarrow \dfrac{2}{3} \lt 1 \le 3 & \mathbf{y = 1} \\ & && \Rightarrow \dfrac{2}{3} \lt 2 \le 3 & \mathbf{y = 2} \\\\ \mathbf{x=4}: & \dfrac{2}{4} \lt &y& \le \dfrac{2}{4} + \dfrac{\sqrt{ 4+2\cdot 4^2 }}{4} \\\\ & \dfrac{1}{2} \lt &y& \le \dfrac{1}{2} + \dfrac{ \sqrt{ 36 }}{4} \\ & \dfrac{1}{2} \lt &y& \le \dfrac{1}{2} + 1.5 \\ & \dfrac{1}{2} \lt &y& \le 2 \\ & && \Rightarrow \dfrac{1}{2} \lt 1 \le 2 & \mathbf{y = 1} \\ & && \Rightarrow \dfrac{1}{2} \lt 2 \le 2 & \mathbf{y = 2} \\\\ \mathbf{x\gt 4}:\\ \lim \limits_{x\to \infty} \frac{2}{x} + \dfrac{\sqrt{ 4+2x^2 }}{x} \\\\ = \lim \limits_{x\to \infty} \frac{2}{x} + \sqrt{ \frac{4}{x^2}+2 } = \sqrt{2} \\ &&& \Rightarrow & \mathbf{y = 2} \\ &&& \Rightarrow & \mathbf{y = 1} \\ \hline \end{array}\)

 

Solution x,y,z:

\(\begin{array}{|r|r|r|c|} \hline x & y\ge x & z=\dfrac{2(x+y)}{xy-2} & \text{solution} \\ \hline 1 & 3 & 8 & \checkmark \\ & 4 & 5 & \checkmark \\ \hline 2 & 2 & 4 & \checkmark \\ \hline & & \mathbf{\text{sum }z = 17} \\ \hline \end{array}\)

 

 

laugh

heureka Sep 25, 2018
 #2
avatar+20025 
+1

Need some help in summing up the following sequence:
4/1.2.3 + 4/2.3.4 + 4/3.4.5 +..............+ 4/998.999.1000

 

Calculate:
\(\begin{array}{|lcll|} \hline \mathbf{ S_n = 4\cdot \left( \dfrac{1}{1 \cdot 2 \cdot 3} + \dfrac{1}{2 \cdot 3 \cdot 4} + \dfrac{1}{3 \cdot 4 \cdot 5} + \dfrac{1}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{1}{n \cdot (n+1) \cdot (n+2)} \right) } \\ \mathbf{ S_n = 4\cdot \sum \limits_{k=1}^n \dfrac 1{k(k+1)(k+2)} } \\ \hline \end{array} \)

 

\(\text{Calculate $ S = \sum \limits_{k=1}^n \dfrac 1{k(k+1)(k+2)} $} \)

 

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 is, we have all black terms left :

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

 

 

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

 

\(\begin{array}{|lcll|} \hline \mathbf{ S_n} & \mathbf{ =} & \mathbf{ 4\cdot \sum \limits_{k=1}^n \dfrac 1{k(k+1)(k+2)} } \\\\ S_n &=& 4\cdot \dfrac{n(n+3)}{4(n+1)(n+2)} \\\\ \mathbf{ S_n} &\mathbf{=} & \mathbf{ \dfrac{n(n+3)}{(n+1)(n+2)} } \quad & | \quad n = 998 \\\\ S_{998} & {=} & \dfrac{998\cdot(998+3)}{(998+1)(998+2)} \\\\ S_{998} & {=} & \dfrac{998\cdot 1001}{999\cdot 1000} \\\\ S_{998} & {=} & \dfrac{499499}{499500} \\\\ S_{998} & {=} & 0.999\overline{997} \\\\ \hline \end{array}\)

 

laugh

heureka Sep 24, 2018
 #1
avatar+20025 
+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 Sep 19, 2018