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Jan 25, 2019
 #1
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Jan 25, 2019
 #1
avatar+23048 
+8

If I have a list of 5 numbers each of them following another [1,2,3,4,5] and I want to multiply all of them together. 

How do I write (1*2)+(1*3)+(1*4)+(1*5)+(2*3)+(2*4)+(2*5)+(3*4)+(3*5)+(4*5)

But with 4000 numbers in the list instead of 5 without having to write every instance out?

Is it possible to make a rule and calculate it in a common calculator? 

 

Table:

\(\begin{array}{|r|r|r|r|r|r} \hline \text{multiply} & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & \color{red}1 & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} \\ \hline 2 & \color{grey}2 & \color{red}4 & \mathbf{6} & \mathbf{8} & \mathbf{10} \\ \hline 3 & \color{grey}3 & \color{grey}6 & \color{red}9 & \mathbf{12} & \mathbf{15} \\ \hline 4 & \color{grey}4 & \color{grey}8 & \color{grey}12 & \color{red}16 & \mathbf{20} \\ \hline 5 & \color{grey}5 & \color{grey}10 & \color{grey}15 & \color{grey}20 & \color{red}25 \\ \hline sum & \color{blue}15 & \color{blue}30 & \color{blue}45 & \color{blue}60 & \color{blue}75 & \color{blue}\text{arithmetical sequence} \\ \hline \end{array} \)

 

\(\text{Let the sum of all numbers in the table $\mathbf{S} = \color{blue}{15+30+45+60+75} = \color{blue}225 $} \\ \begin{array}{rcll} \text{Let the sum we are looking for } \mathbf{s=} \\ && (1*2)+(1*3)+(1*4)+(1*5)\\ && +(2*3)+(2*4)+(2*5) \\ && +(3*4)+(3*5) \\ && +(4*5) \\ &=& \mathbf{2+3+4+5} \\ && \mathbf{+6+8+10} \\ && \mathbf{+12+15} \\ && \mathbf{+20} \\ &=& 85\\ \end{array} \\ \begin{array}{rcll} \text{Let the sum of all square numbers in the diagonal of the table $\mathbf{q} =$} \color{red}{1+4+9+16+25}=\color{red}55 \\ \end{array} \)

 

So we see the formula: \(\boxed{S = 2s + q}\)  or \(\boxed{s = \dfrac{1}{2}\cdot (S-q) }\)

 

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

\(\begin{array}{|rcll|} \hline a_n &=& a_1+(n-1)d \quad | \quad d=a_1 \\ &=& a_1+(n-1)a_1 \\ &=& a_1+ na_1-a_1 \\ &=& na_1 \quad | \quad a_1=(1+n) \dfrac{n}{2} \\ &=& (1+n) \dfrac{n^2}{2} \\\\ S &=& (a_1 + a_n)\cdot \dfrac{n}{2} \\ &=& \left((1+n) \dfrac{n}{2} +(1+n) \dfrac{n^2}{2} \right)\cdot \dfrac{n}{2} \\ &=& \Big((1+n) +(1+n) n \Big)\cdot \dfrac{n^2}{4} \\ &=& (n^2+2n+1 )\cdot \dfrac{n^2}{4} \\ &=& (n+1)^2\cdot \dfrac{n^2}{4} \\ \mathbf{S} & \mathbf{=} & \mathbf{\left[\dfrac{(n+1)\cdot n}{2}\right]^2} \\ \hline \end{array}\)

 

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

\(\begin{array}{|rcll|} \hline \mathbf{q} & \mathbf{=} & \mathbf{\dfrac{(n+1)\cdot n}{2}\cdot \dfrac{(2n+1)}{3}} \quad | \quad \text{from the formula collection} \\ \hline \end{array} \)

 

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

\(\begin{array}{|rcll|} \hline \mathbf{s} & \mathbf{=} & \mathbf{ \dfrac{1}{2}\cdot (S-q) } \\ &=& \dfrac{1}{2}\cdot \left(\left[\dfrac{(n+1)\cdot n}{2}\right]^2 - \dfrac{(n+1)\cdot n}{2}\cdot \dfrac{(2n+1)}{3} \right) \\ &=& \dfrac{(n+1)\cdot n}{4}\cdot \left[ \dfrac{(n+1)\cdot n}{2} - \dfrac{(2n+1)}{3} \right] \\ &=& \dfrac{(n+1)\cdot n}{24}\cdot \Big[ 3n(n+1) - 2(2n+1) \Big] \\ &=& \dfrac{(n+1)\cdot n\cdot ( 3n^2-n-2)}{24} \\ &=& \dfrac{(n+1)\cdot n\cdot (3n+2)\cdot (n-1)}{24} \\\\ \mathbf{s} & \mathbf{=} & \mathbf{\dfrac{(n-1)\cdot n \cdot (n+1)\cdot (3n+2)}{24} } \\ \hline \end{array}\)

 

Example: \([1,2,3,4,5]\)

So \(n = 5 \)

\(\begin{array}{|rcll|} \hline \mathbf{s} & \mathbf{=} & \mathbf{\dfrac{(5-1)\cdot 5 \cdot (5+1)\cdot (3\cdot 5+2)}{24} } \\ &=& \dfrac{4\cdot 5 \cdot 6\cdot 17}{24} \\ &=& 5 \cdot 17 \\ &\mathbf{=}& \mathbf{85}\ \checkmark\\ \hline \end{array}\)

 

Example:\( [1,2,3,4,5, \ldots ,4000]\)

So \(n = 4000\)

\(\begin{array}{|rcll|} \hline \mathbf{s} & \mathbf{=} & \mathbf{\dfrac{(4000-1)\cdot 4000 \cdot (4000+1)\cdot (3\cdot 4000+2)}{24} } \\ &=& \dfrac{3999\cdot 4000 \cdot 4001\cdot 12002 }{24} \\ &=& 1333\cdot 1000 \cdot 4001\cdot 6001 \\ &\mathbf{=}& \mathbf{32\ 005\ 331\ 333\ 000} \\ \hline \end{array} \)

 

 

laugh

Jan 25, 2019
 #3
avatar+103676 
+2

I do not believe the question can be answered. 3.5 is too small there is no solution.

I think that 3.5 was an approximation, not an exact value.  The exact value that they wanted was sqrt(12.8) which is approx 3.5776.  

This is an approximation of the smallest possible perimeter of the octagon.

 

 

Ok now with the proper explanaton, But I am replacing 3.5 with  P which will eventually be replaced with sqrt(12.8)

 

I discovered lots about rotational symmetry while I was answering this question. It is really cool!

 

 

 

For starters why don't I say that the perimeter of the octagon is P units and work from there.  

 

\(a+2b=1\\ a=1-2b\\~\\ 8\sqrt{a^2+b^2}=P\\ \sqrt{1+4b^2-4b+b^2}=\frac{P}{8}\\ \sqrt{5b^2-4b+1}=\frac{P}{8}\\ 5b^2-4b+1=\frac{P^2}{64}\\ 5b^2-4b+\frac{64-P^2}{64}=0\\ \triangle=16-\frac{20(64-P^2)}{64}\\ \triangle=\frac{16*64-20(64-P^2)}{64}\\ \triangle=\frac{4*64-5(64-P^2)}{16}\\ \triangle=\frac{-64+5P^2}{16}\\\)

 

There can be no real solution if the discriminant is negative so 5P^2 must be bigger or equal to 64

\(5P^2\ge64\\ P^2\ge12.8\\ P\ge \sqrt{12.8} \quad \text{which is approximately 3.5777} \)

 

So I am going to assume that the question intended the perimeter of the octagon to be  sqrt(12.8)

which makes the discriminant 0

 

\(5b^2-4b+\frac{64-P^2}{64}=0\\ P=\sqrt{12.8}\\ b=\frac{4\pm0}{10}=\frac{2}{5}\\ a=1-2b=\frac{1}{5}\\ P=8\sqrt{\frac{1}{25}+\frac{4}{25}}=\sqrt{\frac{64*5}{25}}=\sqrt{12.8}\;\;\;\text{Excellent} \)

 

Area of the octogon = \(1- 4 * 0.5* a*b = 1-2*0.2*0.4 = 1-0.16 = 0.84 units^2\)

.
Jan 25, 2019
 #1
avatar+23048 
+8

Let $A_1 A_2 \dotsb A_{11}$ be a regular 11-gon inscribed in a circle of radius 2.

Let $P$ be a point, such that the distance from $P$ to the center of the circle is 3.

Find \[PA_1^2 + PA_2^2 + \dots + PA_{11}^2.\]

 

\(\text{Let $A_n=\dbinom{x_a}{y_a}$, $~x_a =2\cos\Big((n-1)\cdot \dfrac{360^{\circ}}{11} \Big)$, $~y_a =2\sin\Big((n-1)\cdot \dfrac{360^{\circ}}{11} \Big) $, $~ n=1,2,\ldots 11$ }\\ \text{Let $P=\dbinom{x_p}{y_p}$, $~x_p =3\cos(\beta)$, $~y_p =3\sin(\beta)$ } \\ \text{Let $PA^2 = (x_p-x_a)^2 + (y_p-y_a)^2$}\)

 

\(\begin{array}{|rcll|} \hline PA_1^2 &=& \left[ 3\cos(\beta) -2\cos\left(0\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 + \left[ 3\sin(\beta) -2\sin\left(0\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 \\ &=& 13 - 12\left[ \cos(\beta)\cos\left(0\cdot \dfrac{360^{\circ}}{11} \right) + \sin(\beta)\sin\left(0\cdot \dfrac{360^{\circ}}{11} \right)\right]\\\\ PA_2^2 &=& \left[ 3\cos(\beta) -2\cos\left(1\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 + \left[ 3\sin(\beta) -2\sin\left(1\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 \\ &=& 13 - 12\left[ \cos(\beta)\cos\left(1\cdot \dfrac{360^{\circ}}{11} \right) + \sin(\beta)\sin\left(1\cdot \dfrac{360^{\circ}}{11} \right)\right]\\\\ PA_3^2 &=& \left[ 3\cos(\beta) -2\cos\left(2\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 + \left[ 3\sin(\beta) -2\sin\left(2\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 \\ &=& 13 - 12\left[ \cos(\beta)\cos\left(2\cdot \dfrac{360^{\circ}}{11} \right) + \sin(\beta)\sin\left(2\cdot \dfrac{360^{\circ}}{11} \right)\right]\\\\ \ldots \\\\ PA_{11}^2 &=& \left[ 3\cos(\beta) -2\cos\left(10\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 + \left[ 3\sin(\beta) -2\sin\left(10\cdot \dfrac{360^{\circ}}{11} \right) \right]^2 \\ &=& 13 - 12\left[ \cos(\beta)\cos\left(10\cdot \dfrac{360^{\circ}}{11} \right) + \sin(\beta)\sin\left(10\cdot \dfrac{360^{\circ}}{11} \right)\right]\\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline && PA_1^2 + PA_2^2 + PA_3^2 + \dots + PA_{11}^2 \\\\ &=& 11\cdot 13 \\ && -12\cdot \Big\{ \cos(\beta) \Big(\underbrace{ \cos(0\cdot \dfrac{360^{\circ}}{11}) + \cos(1\cdot \dfrac{360^{\circ}}{11}) + \cos(2\cdot \dfrac{360^{\circ}}{11}) + \ldots + \cos(10\cdot \dfrac{360^{\circ}}{11}) }_{=0} \Big) \\ && +\sin(\beta) \Big(\underbrace{ \sin(0\cdot \dfrac{360^{\circ}}{11}) + \sin(1\cdot \dfrac{360^{\circ}}{11}) + \sin(2\cdot \dfrac{360^{\circ}}{11}) + \ldots + \sin(10\cdot \dfrac{360^{\circ}}{11}) \Big) }_{=0} \Big\} \\\\ &=& 11\cdot 13 -12\cdot \Big( \cos(\beta)\cdot 0 +\sin(\beta)\cdot 0 \Big) \\ &=& 11\cdot 13 -12\cdot 0 \\ &=& 11\cdot 13 \\ &\mathbf{=}& \mathbf{143} \\ \hline \end{array}\)

 

\(PA_1^2 + PA_2^2 + \dots + PA_{11}^2 = \mathbf{143}\)

 

laugh

Jan 25, 2019
 #1
avatar+700 
+1
 #2
avatar+5776 
+1
Jan 25, 2019

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