heureka

Let \(x\), \(y\), and \(z\) be real numbers such that \(x^2 + y^2 + z^2 = 1\).

Find the maximum value of \(x + y + z\).

The Cauchy–Schwarz inequality states that for all vectors \( {\displaystyle u} \) and  \({\displaystyle v}\) of an inner product space it is true that
\({\displaystyle |\langle \mathbf {u} ,\mathbf {v} \rangle |^{2}\leq \langle \mathbf {u} ,\mathbf {u} \rangle \cdot \langle \mathbf {v} ,\mathbf {v} \rangle ,}\)

where \({\displaystyle \langle \cdot ,\cdot \rangle } \) is the inner product.

Thank you, CPhill !

Thank you, CPhill !

The cost of renting a bus for a field trip was divided evenly among 20 students.

At the last minute, 10 more students joined the trip.

The cost of renting the bus was then redistributed and evenly split among all of the students.

The cost for each of the original 20 students decreased by $1.50.

What was the total cost of renting the bus, in dollars?

\(\text{Let total cost $=c$} \\ \text{Let students $=s$} \\ \text{Let costs per student $c(s)=\dfrac{c}{s}$} \)

\(\begin{array}{|rcll|} \hline c(30) &=& c(20) - $1.50 \\\\ \dfrac{c}{30} &=& \dfrac{c}{20}- $1.50 \quad | \quad \cdot 30\\\\ c &=& 1.5c - $1.50\cdot 30 \\ c &=& 1.5c - $45 \\ 1.5c-c &=& $45 \\ 0.5c &=& $45 \\ c &=&\dfrac{ $45}{0.5} \\ \mathbf{c} &=& \mathbf{$90} \\ \hline \end{array}\)

The total cost of renting the bus, in dollars was $90

What is the sum of the numbers in row 2019?

\(\begin{array}{|lccccccc|} \hline \text{First row}: & & & &1 \\ \text{Second row}: & & &2 &, &3 \\ \text{Third Row}: & &4 &, &5 &, &6 \\ \text{Fourth row}: & 7 &, &8 &, &9 &, &10 \\ \hline \end{array}\)

\(\begin{array}{|l|ccccccccc|lcr|} \hline \text{row}& & & & & & & & & & &&\text{sum} \\ \hline 1 & & & & &1 & & & & & &=&1 \\ 2 & & & &\color{red}2 &, &\color{green}3 & & & & \left(\dfrac{2+3}{2}\right) \cdot 2 &=& 5 \\ 3 & & &\color{red}4 &, &5 &, &\color{green}6 & & & \left(\dfrac{4+6}{2}\right) \cdot 3 &=& 15 \\ 4 & & \color{red}7 &, &8 &, &9 &, &\color{green}10 & & \left(\dfrac{7+10}{2}\right) \cdot 4 &=& 34 \\ 5 & \color{red}11& , &12 &, &13 &, &14 &, &\color{green}15 & \left(\dfrac{11+15}{2}\right) \cdot 5 &=& 65 \\ \vdots & & & & &\vdots & & & & & \\ 2019 & & & & & a,\ldots ,b & & & & & \left(\dfrac{a+b}{2}\right) \cdot 2019 &=& \ ? \\ \hline n & & & & & a_n,\ldots ,b_n & & & & & \left(\dfrac{a_n+b_n}{2}\right) \cdot n &=& s_n \\ \hline \end{array}\)

\(\mathbf{a_n=\ ?}\)
\(\begin{array}{|lcccccccccc|} \hline \text{First row}: & d_1=1 & &2 & &4 & &7 & & 11 & \ldots \\ \text{Second Row}: & &d_2=1 & &2 & &3 & &4 & \ldots \\ \text{Third row}: & & &d_3=1 & &1 & & 1& \ldots \\ \hline \end{array}\)

\(\begin{array}{lcl} a_n &=& \dbinom{n-1}{0}\cdot d_1 + \dbinom{n-1}{1}\cdot d_2 + \dbinom{n-2}{2}\cdot d_3 \\\\ &=& \dbinom{n-1}{0}\cdot 1 + \dbinom{n-1}{1}\cdot 1 + \dbinom{n-2}{2}\cdot 1 \\\\ &=& 1 + (n-1)\cdot 1 + \dfrac{(n-2)(n-1)}{2\cdot 1}\cdot 1 \\\\ &=& 1 + n-1 + \dfrac{(n-2)(n-1)}{2} \\\\ \mathbf{a_n} &=& \mathbf{n +\dfrac{(n-2)(n-1)}{2}} \\ \end{array}\)

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

\(\begin{array}{|lcccccccccc|} \hline \text{First row}: & d_1=1 & &3 & &6 & &10 & & 15 & \ldots \\ \text{Second Row}: & &d_2=2 & &3 & &4 & &5 & \ldots \\ \text{Third row}: & & &d_3=1 & &1 & & 1& \ldots \\ \hline \end{array}\)

\(\begin{array}{lcl} b_n &=& \dbinom{n-1}{0}\cdot d_1 + \dbinom{n-1}{1}\cdot d_2 + \dbinom{n-2}{2}\cdot d_3 \\\\ &=& \dbinom{n-1}{0}\cdot 1 + \dbinom{n-1}{1}\cdot 2 + \dbinom{n-2}{2}\cdot 1 \\\\ &=& 1 + (n-1)\cdot 2 + \dfrac{(n-2)(n-1)}{2\cdot 1}\cdot 1 \\\\ &=& 1 + 2n-2 + \dfrac{(n-2)(n-1)}{2} \\\\ \mathbf{b_n} &=& \mathbf{2n-1 +\dfrac{(n-2)(n-1)}{2}} \\ \end{array}\)

\(\begin{array}{|rcll|} \hline s_n &=& \left(\dfrac{a_n+b_n}{2}\right) \cdot n \\\\ &=& \left(\dfrac{n +\dfrac{(n-2)(n-1)}{2}+2n-1 +\dfrac{(n-2)(n-1)}{2}}{2}\right) \cdot n \\\\ &=& \left(\dfrac{3n-1 + (n-2)(n-1)}{2} \right) \cdot n \\\\ &=& \left(\dfrac{3n-1 + n^2-3n+2}{2} \right) \cdot n \\\\ \mathbf{s_n} &=& \mathbf{\left(\dfrac{ n^2+1}{2} \right) \cdot n} \\ \hline \end{array}\)

The sum of the numbers in row 2019

\(\begin{array}{|rcll|} \hline \mathbf{s_n} &=& \mathbf{\left(\dfrac{ n^2+1}{2} \right) \cdot n} \\\\ s_{2019} &=& \left(\dfrac{ 2019^2+1}{2} \right) \cdot 2019 \\\\ s_{2019} &=& 2038181 \cdot 2019 \\\\ \mathbf{s_{2019}} &=& \mathbf{4115087439} \\ \hline \end{array}\)

What is the sum of the following sequence:
\(\mathbf{1\cdot 2\cdot 3\cdot 4\cdot 5 + 2\cdot 3\cdot 4\cdot 5\cdot 6 + 3\cdot 4\cdot 5\cdot 6\cdot 7 + 4\cdot 5\cdot 6\cdot 7\cdot 8 +\ldots + 996\cdot 997\cdot 998\cdot 999\cdot 1000}\) .

Using the hockey stick identity and the  binomial coefficient  \(\dbinom nk\)

\(\begin{array}{|rcll|} \hline &&\mathbf{ 1\cdot 2\cdot 3\cdot 4\cdot 5 + 2\cdot 3\cdot 4\cdot 5\cdot 6 + 3\cdot 4\cdot 5\cdot 6\cdot 7 + 4\cdot 5\cdot 6\cdot 7\cdot 8 +\ldots + 996\cdot 997\cdot 998\cdot 999\cdot 1000 } \\\\ &=& \dfrac{5!}{0!} + \dfrac{6!}{1!} + \dfrac{7!}{2!} + \dfrac{8!}{3!} + \dfrac{9!}{4!} + \dfrac{10!}{5!} + \dfrac{11!}{6!} +\ldots + \dfrac{1000!}{995!} \\\\ &=& \dfrac{5!}{5!0!}5! + \dfrac{6!}{5!1!}5! + \dfrac{7!}{5!2!}5! + \dfrac{8!}{5!3!}5! + \dfrac{9!}{5!4!}5! + \dfrac{10!}{5!5!}5! + \dfrac{11!}{5!6!}5! +\ldots + \dfrac{1000!}{5!995!}5! \\\\ &=& \dbinom 55 5! + \dbinom 65 5! + \dbinom 75 5! + \dbinom 85 5! + \dbinom 95 5! + \dbinom {10}5 5! + \dbinom {11}5 5! +\ldots + \dbinom {1000}5 5! \\\\ &=& 5! \left( \underbrace{\dbinom 55 + \dbinom 65 + \dbinom 75 + \dbinom 85 + \dbinom 95 + \dbinom {10}5 + \dbinom {11}5 +\ldots + \dbinom {1000}5}_{\text{long stick of the hockey in Pascals triangle}} \right) \\\\ &=& 5!*\left(\underbrace{\dbinom{1001}{6}}_{\text{bend of the hockey stick(hockey stick identity)} } \right) \\\\ &=& 5!* \dbinom{1001}{6} \\\\ &=& 5!* \dfrac{1001!}{6!995!} \\\\ &=& 120*1376423590241700 \\ &=& \mathbf{165170830829004000} \\ \hline \end{array}\)

2) 

In a certain sequence the first term is \(a_1= 2007\) and the second term is \(a_2 = 2008\).

Furthermore, the values of the remaining terms are chosen so that \( a_n + a_{n + 1} + a_{n + 2} = n\) for all \(n \ge 1\).

Determine \(a_{1000}\).

\(\begin{array}{|lrcll|} \hline (1) & a_n + a_{n + 1} + a_{n + 2} &=& n \\ (2) & a_{n+1} + a_{n + 2} + a_{n + 3} &=& n+1 \\ \hline (2)-(1): & a_{n+1} + a_{n + 2} + a_{n + 3}-(a_n + a_{n + 1} + a_{n + 2}) &=& n+1-n \\ & a_{n+1} + a_{n + 2} + a_{n + 3}-a_n - a_{n + 1} - a_{n + 2} &=& 1 \\ & a_{n + 3}-a_n &=& 1 \\ & \mathbf{a_{n + 3}} &=& \mathbf{1+a_n} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline & \mathbf{a_{n + 3}} &=& \mathbf{1+a_n} \\ \hline n= 1: & a_{1 + 3}=\mathbf{a_{4}} &=& \mathbf{1+a_1} \\ \hline n= 4: & a_{4 + 3}=a_{7} &=& 1+a_4 \\ & &=& 1+( 1+a_1) \\ & \mathbf{ a_{7}} &=& \mathbf{2+a_1} \\ \hline n= 7: & a_{7 + 3}=a_{10} &=& 1+a_7 \\ & &=& 1+( 2+a_1) \\ & \mathbf{ a_{10}} &=& \mathbf{3+a_1} \\ \hline n= 10: & a_{10 + 3}=a_{13} &=& 1+a_{10} \\ & &=& 1+( 3+a_1) \\ & \mathbf{ a_{13}} &=& \mathbf{4+a_1} \\ \hline \ldots \\ \hline \end{array}\)

generalized:

\(\begin{array}{|rcll|} \hline \mathbf{a_{4}} &=& \mathbf{1+a_1} \\ \mathbf{ a_{7}} &=& \mathbf{2+a_1} \\ \mathbf{ a_{10}} &=& \mathbf{3+a_1} \\ \mathbf{ a_{13}} &=& \mathbf{4+a_1} \\ \ldots \\ \mathbf{ a_{1+3m}} &=& \mathbf{m+a_1} \\ \hline \end{array} \)

\(\mathbf{a_{1000} =\ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{ a_{1+3m}} &=& \mathbf{m+a_1} \\ \hline 1+3m &=& 1000 \\ 3m &=& 999 \\ m &=& 333 \\ a_{1000} &=& 333+a_1 \quad | \quad a_1 =2007 \\ a_{1000} &=& 333 + 2007 \\ \mathbf{a_{1000}} &=& \mathbf{2340} \\ \hline \end{array}\)

1)
Find the sum of the infinite series
\(1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots\).

\(\text{Let $r=\dfrac{1}{1998} $}\)

\(\begin{array}{|rcll|} \hline s&=& \mathbf{1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots} \\ s&=& \mathbf{1+2r +3r^2+4r^3+\cdots + nr^{n-1} + \ldots} \\ \hline \end{array} \)

\(\begin{array}{|rcrl|} \hline s_n &=& 1+&2r +3r^2+4r^3+\cdots + nr^{n-1} \quad |\quad \cdot r \\ rs_n &=& & 1r +2r^2+3r^3+\cdots + (n-1)r^{n-1}+nr^n \\ \hline s_n-rs_n &=& 1+&r+r^2+r^3+\ldots + r^{n-1} -nr^n \\ \hline \end{array}\\ \begin{array}{|rclrcrl|} \hline s_n(1-r) &=& \underbrace{1+r+r^2+r^3+\ldots + r^{n-1}}_{=S_n(\text{ geometric progression})} -nr^n \\ s_n(1-r) &=& S_n -nr^n & S_n &=& 1+&r+r^2+r^3+\ldots + r^{n-1} \quad | \quad \cdot r\\ & & & rS_n &=& &r+r^2+r^3+\ldots + r^{n-1}+r^n \\ \hline & & & S_n-rS_n &=& 1- &r^n \\ & & & S_n(1-r) &=& 1- &r^n \\ & & & S_n &=& \dfrac{1- r^n}{1-r} \\ s_n(1-r) &=& \dfrac{1- r^n}{1-r} -nr^n \\ s_n &=& \dfrac{1- r^n}{(1-r)^2} - \dfrac{nr^n}{1-r} \\ s &=& \lim \limits_{n\to \infty} s_n \\ s &=& \lim \limits_{n\to \infty} \dfrac{1- r^n}{(1-r)^2} - \dfrac{nr^n}{1-r} \\ && \boxed{\lim \limits_{n\to \infty} r^n = \lim \limits_{n\to \infty} \left(\dfrac{1}{1998}\right)^n = 0} \\ s &=& \dfrac{1- 0}{(1-r)^2} - 0 \\ s &=& \dfrac{1}{(1-r)^2} \\ s &=& \dfrac{1}{\left(1-\dfrac{1}{1998}\right)^2} \\ s &=& \left(\dfrac{1998}{1997}\right)^2 \\ \mathbf{s} &=& \mathbf{1.00100175300507095\ldots...} \\ \hline \end{array}\)

What is the sum of the following sequence:
\(1\cdot 2\cdot 3\cdot 4\cdot 5 + 2\cdot 3\cdot 4\cdot 5\cdot 6 + 3\cdot 4\cdot 5\cdot 6\cdot 7 + 4\cdot 5\cdot 6\cdot 7\cdot 8 +\ldots + 996\cdot 997\cdot 998\cdot 999\cdot 1000\) .

Using the hockey stick identity:

\(\begin{array}{|rcll|} \hline &&\mathbf{ 1\cdot 2\cdot 3\cdot 4\cdot 5 + 2\cdot 3\cdot 4\cdot 5\cdot 6 + 3\cdot 4\cdot 5\cdot 6\cdot 7 + 4\cdot 5\cdot 6\cdot 7\cdot 8 +\ldots + 996\cdot 997\cdot 998\cdot 999\cdot 1000 } \\\\ &=& 120 + 720 + 2510 + 6720 + 15120 + 30240 + 55440 +\ldots + 990034950024000 \\ &=& 120 +6*120 + 21*120 + 56*120 + 126*120 + 252*120 + 462*120 +\ldots + 8250291250200*120 \\ &=& 120*(\underbrace{1+ 6 + 21 + 56 + 126 + 252 + 462 +\ldots + 8250291250200}_{\text{long stick of the hockey in Pascals triangle}} ) \\ &=& 120*\left(\dbinom{5}{5}+\dbinom{6}{5}+\dbinom{7}{5}+\dbinom{8}{5}+\dbinom{9}{5}+\dbinom{10}{5}+\dbinom{11}{5}+\ldots +\dbinom{1000}{5} \right) \\ &=& 120*\left(\underbrace{\dbinom{1001}{6}}_{\text{bend of the hockey stick(hockey stick identity)} } \right) \\ &=& 120* \dbinom{1001}{6} \\ &=& 120*1376423590241700 \\ &=& \mathbf{165170830829004000} \\ \hline \end{array} \)

If E is the middle point of the side CA of the triangle ABC and
if R is the area of the triangle, prove that:

\(\cot(AEB) = \dfrac{ BC^2 - BA^2 } { 4R } \)

\(\text{Let $\angle AEB= \epsilon$} \\ \text{Let $AE=EC$} \)

\(\begin{array}{|rcll|} \hline BC^2 &=& AE^2 + BE^2 - 2\cdot AE\cdot BE \cdot \cos(180^\circ-\epsilon ) \\ &=& AE^2 + BE^2 + 2\cdot AE\cdot BE \cdot \cos(\epsilon ) \\\\ BA^2 &=& EC^2 + BE^2 - 2\cdot EC\cdot BE \cdot \cos(\epsilon ) \\\\ BC^2 - BA^2 &=& 4\cdot AE\cdot BE \cdot \cos(\epsilon ) \quad & | \quad EC=AE \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline R &=& R_1 + R_2 \quad & | \quad R_1 = \text{area}~\triangle AEB,\quad R_2 = \text{area}~\triangle ECB \\ &=& \dfrac{AE\cdot BE\cdot \sin(\epsilon) }{2} + \dfrac{EC\cdot BE \sin(180^\circ-\epsilon) }{2} \quad & | \quad EC=AE \\ &=& \dfrac{AE\cdot BE\cdot \sin(\epsilon) }{2} + \dfrac{AE\cdot BE \sin(\epsilon) }{2} \\ &=& AE\cdot BE\cdot \sin(\epsilon) \\\\ AE\cdot BE &=& \dfrac{R}{\sin(\epsilon)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline BC^2 - BA^2 &=& 4\cdot AE\cdot BE \cdot \cos(\epsilon ) \quad | \quad AE\cdot BE = \dfrac{R}{\sin(\epsilon)} \\ BC^2 - BA^2 &=& 4\cdot \dfrac{R}{\sin(\epsilon)} \cdot \cos(\epsilon ) \\ BC^2 - BA^2 &=& 4R \cdot \cot(\epsilon ) \\\\ \mathbf{\cot(\epsilon )} &=& \mathbf{\dfrac{BC^2 - BA^2}{4R}} \\ \hline \end{array} \)