heureka

How many interior diagonals does an icosahedron have?

(An interior diagonal is a segment connecting two vertices which do not lie on a common face.)

I assume:

\(\dbinom{\text{vertices}}{2} - \text{edges} \quad | \quad \text{vertices}=12,\ \text{edges} = 30\)

\(\begin{array}{|rcll|} \hline && \mathbf{\dbinom{\text{vertices}}{2} - \text{edges}} \quad | \quad \text{vertices}=12,\ \text{edges} = 30 \\\\ &=& \dbinom{\text{12}}{2} - 30 \\\\ &=& \dfrac{12}{2}*\dfrac{11}{1} - 30 \\\\ &=& 66 - 30 \\ &=& \mathbf{36} \\ \hline \end{array}\)

2) Evaluate the series

\(\dfrac{1}{3^1}+\dfrac{4}{3^2}+\dfrac{7}{3^3}+\dfrac{10}{3^4}+\cdots \).

\(\begin{array}{|rcll|} \hline (1+0*3)\left(1*\frac{1}{3}\right) +(1+1*3)\left(1*\frac{1}{3^2}\right) +(1+2*3)\left(1*\frac{1}{3^3}\right) +(1+3*3)\left(1*\frac{1}{3^4}\right) + \dotsb \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \text{arithmetical sequence:} \\ {\color{red}1}+({\color{red}1}+1*{\color{orange}3})+({\color{red}1}+2*{\color{orange}3})+({\color{red}1}+3*{\color{orange}3})+\dotsb+\Big({\color{red}1}+(n-1)*{\color{orange}3}\Big)+\dotsb \\ \boxed{a={\color{red}1},\ d={\color{orange}3} \text{ is the common difference} } \\\\ \text{geometric series:} \\ {\color{blue}\frac{1}{3} } +{\color{blue}\frac{1}{3} }\left({\color{green}\frac{1}{3}}\right)^1 +{\color{blue}\frac{1}{3} }\left({\color{green}\frac{1}{3}}\right)^2 +{\color{blue}\frac{1}{3} }\left({\color{green}\frac{1}{3}}\right)^3 +\dotsb +{\color{blue}\frac{1}{3} }\left({\color{green}\frac{1}{3}}\right)^{n-1} +\dotsb\\ \boxed{ b={\color{blue}\frac{1}{3} },\ r={\color{green}\frac{1}{3}}\ \text{ is the common ratio} } \\ \hline \end{array}\)

Formula:
sum of a infinite arithmetico-geometric sequence
\(\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{\left(\dfrac{b}{1-r}\right) \Bigg( a+d\left( \dfrac{r}{1-r} \right) \Bigg)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{\left(\dfrac{b}{1-r}\right) \Bigg( a+d\left( \dfrac{r}{1-r} \right) \Bigg)} \\\\ && \boxed{a={\color{red}1},\ d={\color{orange}3} \text{ is the common difference} } \\ &&\boxed{ b={\color{blue}\frac{1}{3} },\ r={\color{green}\frac{1}{3}}\ \text{ is the common ratio} } \\\\ s &=& \left(\dfrac{{\color{blue}\frac{1}{3}} }{1-{ \color{green}\frac{1}{3}} }\right) \Bigg( {\color{red}1} + { \color{orange}3} \left( \dfrac{ {\color{green}\frac{1}{3} }}{1-{\color{green}\frac{1}{3}} } \right) \Bigg) \\\\ s &=& \frac{1}{3}*\frac{3}{2} \Bigg( {\color{red}1} + \dfrac{1}{1-{\color{green}\frac{1}{3}} } \Bigg) \\\\ s &=& \frac{1}{2} *\frac{5}{2} \\\\ \mathbf{s} &=& \mathbf{\frac{5}{4}} \\ \hline \end{array}\)

\(\mathbf{\dfrac{1}{3^1}+\dfrac{4}{3^2}+\dfrac{7}{3^3}+\dfrac{10}{3^4}+\cdots} = \mathbf{\dfrac{5}{4}}\)

1) Find
\(\dfrac{1}{4} + \dfrac{1}{4^2} + \dfrac{1}{4^3} + \dotsb\) .

Geometric progression: \( a,\ ar,\ ar^2,\ ar^3,\ ar^4,\ \ldots\)

\(\begin{array}{|rcll|} \hline \mathbf{\text{Geometric progression: } a,\ ar,\ ar^2,\ ar^3,\ ar^4,\ \ldots,\qquad a=\dfrac{1}{4},\ r=\dfrac{1}{4} } \\ \hline \dfrac{1}{4},\ \dfrac{1}{4}*\left(\dfrac{1}{4}\right)^1,\ \dfrac{1}{4}\left(\dfrac{1}{4}\right)^2,\ \dfrac{1}{4}\left(\dfrac{1}{4}\right)^3,\ \dfrac{1}{4}\left(\dfrac{1}{4}\right)^4,\ \ldots \\ \hline \end{array}\)

Infinite geometric series, the sum formula: \(\dfrac{a}{1-r}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{a}{1-r}} &=& \dfrac{\frac{1}{4}}{1-\frac{1}{4}} \quad |\quad a=\dfrac{1}{4},\ r=\dfrac{1}{4} \\\\ &=& \dfrac{4}{4*3} \\\\ &=& \mathbf{\dfrac{1}{3}} \\ \hline \end{array}\)

\(\mathbf{\dfrac{1}{4} + \dfrac{1}{4^2} + \dfrac{1}{4^3} + \dotsb} = \mathbf{\dfrac{1}{3}}\)

Square MATH has sides of length 4, and N is the midpoint of TH.
A circle with radius 2 and center N intersects a circle with radius 4 and center M at points P and H.
What is the distance from P to MH.

\(\begin{array}{|rcll|} \hline \mathbf{\text{Circle with }r=4 \\ \text{ and center } M(0,4):} \\ \hline x^2+(y-4)^2 &=& 4^2 \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\text{Circle with }r=2 \\ \text{ and center } N(2,0):} \\ \hline (x-2)^2+y^2 &=& 2^2 \\ \hline \end{array} \)

\(\begin{array}{|lcll|} \hline \mathbf{\text{Intersects at }P(x_P,\ y_P):} \\ \begin{array}{|rcll|} \hline x_P^2+(y_P-4)^2 &=& 16 \\ x_P^2+y_P^2-8yP + 16 &=& 16 \\ x_P^2+y_P^2-8yP &=& 0 \\ \mathbf{x_P^2+y_P^2} &=& \mathbf{8y_P} \\ \hline \end{array} \begin{array}{|rcll|} \hline (x_P-2)^2+y_P^2 &=& 4 \\ x_P^2-4x_P + 4 +y_P^2 &=& 4 \\ x_P^2 +y_P^2-4x_P &=& 0 \\ \mathbf{x_P^2 +y_P^2}&=& \mathbf{4x_P} \\ y_P^2 &=& 4x_P -x_P^2 \\ \mathbf{ y_P } &=& \mathbf{\sqrt{ 4x_P -x_P^2}} \\ \hline \end{array} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{x_P^2+y_P^2} &=& \mathbf{8y_P} \quad | \quad \mathbf{x_P^2 +y_P^2=4x_P},\ \mathbf{ y_P =\sqrt{ 4x_P -x_P^2}} \\ 4x_P &=& 8\sqrt{ 4x_P -x_P^2} \quad | \quad :4 \\ x_P &=& 2\sqrt{ 4x_P -x_P^2} \quad | \quad \text{square both sides} \\ x_P^2 &=& 4(4x_P -x_P^2) \\ x_P^2 &=& 16x_P -4x_P^2 \\ 5x_P^2 &=& 16x_P \\ 5x_P^2-16x_P &=& 0 \\ x_P(5x_P-16) &=& 0 \quad | \quad x_P \neq 0~! \\\\ 5x_P-16 &=& 0 \\ 5x_P &=& 16 \\ x_P &=& \dfrac{16}{5} \\ \mathbf{x_P} &=& \mathbf{3.2} \\ \hline \end{array}\)

The distance from P to MH is \(\mathbf{3.2}\)

For what value of c are the three points (-3,1), (5,5), and (6,c) on the same line?

\(\begin{array}{|rcll|} \hline \dfrac{c-5}{6-5} &=& \dfrac{5-1}{5-(-3)} \\\\ c-5 &=& \dfrac{4}{8} \\\\ c-5 &=& \dfrac{1}{2} \\\\ c &=& 5+ \dfrac{1}{2} \\\\ \mathbf{c} &=& \mathbf{5.5} \\ \hline \end{array}\)

For what value of c are the three points (-3,1), (5,5), and (6,c) on the same line?

\(\begin{array}{|rcll|} \hline \dfrac{c-5}{6-5} &=& \dfrac{5-1}{5-(-3)} \\\\ c-5 &=& \dfrac{4}{8} \\\\ c-5 &=& \dfrac{1}{2} \\\\ c &=& 5+ \dfrac{1}{2} \\\\ \mathbf{c} &=& \mathbf{5.5} \\ \hline \end{array}\)

Use differences to find a pattern in the sequence
\(2,\ 12,\ 29,\ 61,\ 116,\ 202,\ 327,\ \dots\)

\(\small{ \begin{array}{lccccccc} & {\color{red}d_0 = 2} && 12 && 29 && 61 && 116 && 202 && 327&& \cdots \\ \text{1. Difference } && {\color{red}d_1 = 10} && 17 && 32 && 55 && 86 && 125&& \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = 7} && 15 && 23 && 31 && 39&& \cdots \\ \text{3. Difference } &&&& {\color{red}d_3 = 8} && 8 && 8&& 8&& \cdots \\ \end{array} }\)

Formula:

\(\boxed{~ \begin{array}{rcl} a_n &=& \binom{n-1}{0}\cdot {\color{red}d_0 } + \binom{n-1}{1}\cdot {\color{red}d_1 } + \binom{n-1}{2}\cdot {\color{red}d_2 } + \binom{n-1}{3}\cdot {\color{red}d_3 } \end{array} ~}\)

\(\begin{array}{rcl} a_n &=& \binom{n-1}{0}\cdot {\color{red} 2 } + \binom{n-1}{1}\cdot {\color{red} 10 } + \binom{n-1}{2}\cdot {\color{red} 7 } + \binom{n-1}{3}\cdot {\color{red} 8 } \\ \hline && \binom{n-1}{0} = 1 \\ && \binom{n-1}{1} = n-1 \\ && \binom{n-1}{2} = ( \frac{n-1}{2} ) \cdot ( \frac{n-2}{1} ) \\ && \binom{n-1}{3} = ( \frac{n-1}{3} ) \cdot ( \frac{n-2}{2} )\cdot ( \frac{n-3}{1} ) \\ \hline a_n &=& 1* {\color{red} 2 }+ (n-1)\cdot {\color{red}10} + ( \frac{n-1}{2} ) \cdot ( \frac{n-2}{1} )\cdot {\color{red}7} + ( \frac{n-1}{3} ) \cdot ( \frac{n-2}{2} )\cdot ( \frac{n-3}{1} )\cdot {\color{red}8} \quad | \quad \cdot 6\\ 6\cdot a_n &=& 12 + (n-1)\cdot 60 + ( n-1 ) \cdot ( n-2 )\cdot 21 + ( n-1 ) \cdot ( n-2 )\cdot ( n-3 )\cdot 8 \\ 6\cdot a_n &=& 12+ (n-1) \left[~ 60 + ( n-2 )\cdot 21 + ( n-2 )\cdot ( n-3 )\cdot 8 ~\right] \\ 6\cdot a_n &=& 12+(n-1) \left[~ 60 + 21n-42 + (n^2 - 5n + 6)\cdot 8 ~\right] \\ 6\cdot a_n &=& 12+(n-1) \left(~ 18 + 21n + 8n^2 - 40n + 48 ~\right) \\ 6\cdot a_n &=& 12+(n-1) \left(~ 66 - 19n + 8n^2 ~\right) \\ 6\cdot a_n &=& 12+ 66n - 19n^2 + 8n^3 - 66 + 19n - 8n^2 \\ 6\cdot a_n &=& -54 + 85n - 27n^2 + 8n^3 \\\\ a_n &=& \dfrac{ -54 + 85n - 27n^2 + 8n^3 }{6} \\\\ \mathbf{a_n} &=& \mathbf{ -9 + \dfrac{n(85 - 27n + 8n^2) }{6} } \\ \end{array}\)

Find the value of k so that
\(3 + \dfrac{3 + k}{4} + \dfrac{3 + 2k}{4^2} + \dfrac{3 + 3k}{4^3} + \dotsb = 8\).

\(\begin{array}{|rcll|} \hline 3*1 + (3 + k)\left(1*\frac{1}{4}\right) + (3 + 2k)\left(1* \frac{1}{4^2} \right) + (3 + 3k)\left(1* \frac{1}{4^3} \right) + \dotsb + (3 + (n-1)k)\left(1*\left(\frac{1}{4}\right)^{n-1}\right) + \dotsb = 8 \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \text{arithmetical sequence:} \\ {\color{red}3}+({\color{red}3}+k)+({\color{red}3}+2k)+({\color{red}3}+3k)+\dotsb \\ \boxed{a={\color{red}3},\ k \text{ is the common difference} } \\\\ \text{geometric series:} \\ {\color{blue}1}+{\color{blue}1}*{\color{green}\frac{1}{4}}+{\color{blue}1}*({\color{green}\frac{1}{4}})^2+{\color{blue}1}*({\color{green}\frac{1}{4}})^3+\dotsb \\ \boxed{ b={\color{blue}1},\ r={\color{green}\frac{1}{4}}\ \text{ is the common ratio} } \\ \hline \end{array}\)

The sum  of the first n terms of an arithmetico–geometric sequence has the form:

\(\begin{array}{|rcll|} \hline \mathbf{s_n} &=& \mathbf{ \dfrac{ ab-(a+nk)br^n} {1-r} + \dfrac{kbr(1-r^n)} {(1-r)^2}} \\ &&\boxed{a=3,\ b=1,\ r=\frac{1}{4}} \\\\ &=& \dfrac{3*1-(3+nk)*1*\frac{1}{4^n}}{1-\frac{1}{4}} + \frac{ k*1*\frac{1}{4}(1-\frac{1}{4^n})} { (1-\frac{1}{4})^2 } \\\\ &=&\frac{4}{3}\left( 3-\frac{3}{4^n}-\frac{nk}{4^n} \right) +\frac{16}{9}\left( \frac{k}{4}-\frac{k}{4*4^n} \right) \\\\ &=& 4-\frac{1}{4^{n-1}}-\frac{nk}{3*4^{n-1}} + \frac{4}{9}k-\frac{k}{4*4^n}*\frac{4*4}{9} \\\\ &=& 4-\frac{1}{4^{n-1}}-\frac{nk}{3*4^{n-1}} + \frac{4}{9}k-\frac{k}{9*4^{n-1}} \\\\ &=& 4+ \frac{4}{9}k -\frac{1}{4^{n-1}}\left(1+\frac{nk}{3}+\frac{k}{9}\right) \\ &&\boxed{\lim \limits_{n\to\infty} \frac{1}{4^{n-1}}\left(1+\frac{nk}{3}+\frac{k}{9}\right)=0} \\ 8 &=& 4+ \frac{4}{9}k \\ 4 &=& \frac{4}{9}k \quad | \quad : 4 \\ 1 &=& \frac{1}{9}k \quad | \quad * 9 \\ 9 &=& k \\ \mathbf{k} &=& \mathbf{9} \\ \hline \end{array}\)

Formula:

infinite arithmetico–geometric sequence

\(\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{\left(\dfrac{b}{1-r}\right) \Bigg( a+k\left( \dfrac{r}{1-r} \right) \Bigg)} \\ \hline \end{array}\)

The sum of the first \(n\) terms of a certain sequence is
\(n(n + 1)(n + 2)\).
Find the tenth term of the sequence.

\(\begin{array}{|c|l|l|l|l|} \hline n & s_n=n(n+1)(n+2) & \\ \hline 1 & s_1=1*2*3=6 & s_1 = a_1 & 6= a_1 & a_1 = 6 \\ \hline 2 & s_2=2*3*4=24 & s_2 = s_1+a_2 & 24=6+a_2 & a_2 = 18 \\ \hline 3 & s_3=3*4*5=60 & s_3 = s_2+a_3 & 60=24+a_3 & a_3 = 36 \\ \hline 4 & s_4=4*5*6=120 & s_4 = s_3+a_4 & 120=60+a_4 & a_4 = 60 \\ \hline 5 & s_5=5*6*7=210 & s_5 = s_4+a_5 & 210=120+a_5 & a_5 = 90 \\ \hline 6 & s_6=6*7*8=336 & s_6 = s_5+a_6 & 336=210+a_6 & a_6 = 126 \\ \hline 7 & s_7=7*8*9=504 & s_7 = s_6+a_7 & 504=336+a_7 & a_7 = 168 \\ \hline 8 & s_8=8*9*10=720 & s_8 = s_7+a_8 & 720=504+a_8 & a_8 = 216 \\ \hline 9 & s_9=9*10*11=990 & s_9 = s_8+a_9 & 990=720+a_9 & a_9 = 270 \\ \hline 10 & s_{10}=10*11*12=1320 & s_{10} = s_9+a_{10} & 1320=990+a_{10} & \mathbf{a_{10} = 330} \\ \hline \end{array}\)

Simplify
\(\dfrac{1}{1 + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{4}} + \dots + \dfrac{1}{\sqrt{99} + \sqrt{100}}\).

\(\small{ \begin{array}{|rcll|} \hline &&\mathbf{\dfrac{1}{1 + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{4}} + \dots + \dfrac{1}{\sqrt{99} + \sqrt{100}}} \\\\ &=& \dfrac{1 - \sqrt{2}} {(1 + \sqrt{2})(1 - \sqrt{2})} + \dfrac{\sqrt{2} - \sqrt{3}} {(\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3})} + \dfrac{\sqrt{3} - \sqrt{4}} {(\sqrt{3} + \sqrt{4})(\sqrt{3} - \sqrt{4})} + \dots + \dfrac{\sqrt{99} - \sqrt{100}} {(\sqrt{99} + \sqrt{100})(\sqrt{99} - \sqrt{100})} \\\\ &=& \dfrac{1 - \sqrt{2}} {1-2} + \dfrac{\sqrt{2} - \sqrt{3}} {2-3} + \dfrac{\sqrt{3} - \sqrt{4}} {3-4} + \dots + \dfrac{\sqrt{99} - \sqrt{100}} {99-100} \\\\ &=& \dfrac{1 - \sqrt{2}} {-1} + \dfrac{\sqrt{2} - \sqrt{3}} {-1} + \dfrac{\sqrt{3} - \sqrt{4}} {-1} + \dots + \dfrac{\sqrt{99} - \sqrt{100}} {-1} \\\\ &=& -1 +\underbrace{ \sqrt{2} -\sqrt{2} + \sqrt{3} -\sqrt{3} + \sqrt{4} + \dots -\sqrt{99}}_{=0} + \sqrt{100} \\\\ &=& -1 + \sqrt{100} \\\\ &=& -1 + 10 \\\\ &=& \mathbf{ 9 } \\ \hline \end{array} }\)

\(\dfrac{1}{1 + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{4}} + \dots + \dfrac{1}{\sqrt{99} + \sqrt{100}} = \mathbf{ 9}\)