heureka

Find the largest integer that divides all integers of the form  \(7^{2n + 1} + 1\), for integers \(n \ge 0\).

My attempt:

\(\text{if }~n=0:~ 7^{2n + 1} + 1 = 7^{2*0 + 1} + 1 =7 + 1 = 8 \)

The largest integer that divides \(7^{2n + 1} + 1\), if \(n=0\) is \(8\).

The question is: \(7^{2n + 1} + 1 \equiv 0 \pmod{8} \ ?\)

\(\begin{array}{|rcll|} \hline 7^{2n + 1} + 1 &\equiv& 0 \pmod{8} \ ? \\ \hline && 7^{2n + 1} + 1 \pmod{8} \\ &\equiv& 7^{2n}*7 + 1 \pmod{8} \\ &\equiv& \left(7^2 \right)^n*7 + 1 \pmod{8} \\ &\equiv& {49}^n*7 + 1 \pmod{8} \quad |\quad 49 &\equiv& 1 \pmod{8} \\ &\equiv& {1}^n*7 + 1 \pmod{8} \\ &\equiv& 7 + 1 \pmod{8} \\ &\equiv& 8 \pmod{8} \\ &\equiv& 8-8 \pmod{8} \\ &\equiv& 0 \pmod{8} \\ \hline \end{array} \)

Find the largest integer that divides all integers of the form \(7^{2n + 1} + 1\) is 8

Compute

\(\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{100^3 - 100}\).

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{100^3 - 100}} \\\\ &=& \dfrac{1}{2(2^2 - 1)} + \dfrac{1}{3(3^2 - 1)} + \dfrac{1}{4(4^2 - 1)} + \dots + \dfrac{1}{100(100^2 - 1)} \\\\ &=& \dfrac{1}{2(2 - 1)(2+1)} + \dfrac{1}{3(3 - 1)(3+1)} + \dfrac{1}{4(4 - 1)(4+1)} + \dots + \dfrac{1}{100(100 - 1)(100+1)} \\\\ &=& \dfrac{1}{\mathbf{1}*2*3} + \dfrac{1}{\mathbf{2}*3*4} + \dfrac{1}{\mathbf{3}*4*5} + \dots + \dfrac{1}{\mathbf{99}*100*101} \\\\ &=& \sum \limits_{n=1}^{99}\dfrac{1}{n(n+1)(n+2)} \\\\ &=& \sum \limits_{n=1}^{99}\dfrac{1}{n(n+1)} * \dfrac{1}{(n+2)} \quad | \quad \dfrac{1}{n(n+1)}= \dfrac{1}{n}- \dfrac{1}{(n+1)}\\\\ &=& \sum \limits_{n=1}^{99}\left(\dfrac{1}{n}- \dfrac{1}{(n+1)}\right) \dfrac{1}{(n+2)} \\\\ &=& \sum \limits_{n=1}^{99}\left(\dfrac{1}{n(n+2)}- \dfrac{1}{(n+1)(n+2)} \right) \quad | \quad \dfrac{1}{n(n+2)}=\dfrac{1}{2}\left( \dfrac{1}{n}- \dfrac{1}{(n+2)} \right) \\\\ &=& \sum \limits_{n=1}^{99}\Bigg(\dfrac{1}{2}\left( \dfrac{1}{n}- \dfrac{1}{(n+2)} \right)- \dfrac{1}{(n+1)(n+2)} \Bigg) \quad | \quad \dfrac{1}{(n+1)(n+2)}=\left( \dfrac{1}{(n+1)}- \dfrac{1}{(n+2)} \right) \\\\ &=& \sum \limits_{n=1}^{99}\Bigg(\dfrac{1}{2}\left( \dfrac{1}{n}- \dfrac{1}{(n+2)} \right)- \left( \dfrac{1}{(n+1)}- \dfrac{1}{(n+2)} \right) \Bigg) \\\\ &=& \sum \limits_{n=1}^{99}\left(\dfrac{1}{2}*\dfrac{1}{n}- \dfrac{1}{2}*\dfrac{1}{(n+2)} - \dfrac{1}{(n+1)}+ \dfrac{1}{(n+2)} \right) \\\\ &=& \sum \limits_{n=1}^{99}\left(\dfrac{1}{2}*\dfrac{1}{n}- \dfrac{1}{(n+1)} + \dfrac{1}{2}*\dfrac{1}{(n+2)} \right) \\\\ &=& \dfrac{1}{2}\sum \limits_{n=1}^{99}\left(\dfrac{1}{n} \right) -\sum \limits_{n=1}^{99}\left(\dfrac{1}{n+1} \right) +\dfrac{1}{2}\sum \limits_{n=1}^{99}\left(\dfrac{1}{n+2} \right) \\ \\ &=& \dfrac{1}{2}\sum \limits_{n=1}^{99}\left(\dfrac{1}{n} \right) -\sum \limits_{n=2}^{100}\left(\dfrac{1}{n} \right) +\dfrac{1}{2}\sum \limits_{n=3}^{101}\left(\dfrac{1}{n} \right) \\ \\ &=& \dfrac{1}{2}\left(1+\dfrac{1}{2} \right)+ \dfrac{1}{2}\sum \limits_{n=3}^{99}\left(\dfrac{1}{n} \right) -\dfrac{1}{2}-\dfrac{1}{100} -\sum \limits_{n=3}^{99}\left(\dfrac{1}{n} \right) +\dfrac{1}{2}*\dfrac{1}{100}+ \dfrac{1}{2}*\dfrac{1}{101} + \sum \limits_{n=3}^{99}\left(\dfrac{1}{n} \right) \\ \\ &=& \left( \dfrac{1}{2} - 1 + \dfrac{1}{2} \right) \sum \limits_{n=3}^{99}\left(\dfrac{1}{n} \right) +\dfrac{1}{2} + \dfrac{1}{4}-\dfrac{1}{2}-\dfrac{1}{100} +\dfrac{1}{200}+ \dfrac{1}{202} \\ \\ &=& \left( 1 - 1 \right) \sum \limits_{n=3}^{99}\left(\dfrac{1}{n} \right) + \dfrac{1}{4}-\dfrac{1}{100}+\dfrac{1}{200}+ \dfrac{1}{202} \\\\ &=&0 + \dfrac{50}{200}-\dfrac{2}{200}+\dfrac{1}{200}+ \dfrac{1}{202} \\\\ &=& \dfrac{49}{200}+ \dfrac{1}{202} \\\\ &=& \dfrac{49*202+200}{200*202} \\\\ &=& \dfrac{10098}{40400} \\\\ &=& \mathbf{\dfrac{5049}{20200} } \\ \hline \end{array}\)

Let

\(\mathbf{v}_1 = < 5, -1 >\) and \(\mathbf{v}_2 = < -4, 2 >\).
Compute \(\mathbf{v}_1 + \mathbf{v}_2\), \(\mathbf{v}_1 – \mathbf{v}_2\), and \(\mathbf{v}_2 – \mathbf{v}_1\).
\(\mathbf{v}_1 + \mathbf{v}_2 =\)
\(\mathbf{v}_1 – \mathbf{v}_2=\)
\(\mathbf{v}_2 – \mathbf{v}_1 =\)

\(\begin{array}{|rcll|} \hline \mathbf{v}_1 + \mathbf{v}_2 &=& < 5, -1 > + < -4, 2 > \\ \mathbf{v}_1 + \mathbf{v}_2 &=& < 5-4, -1+2 > \\ \mathbf{v}_1 + \mathbf{v}_2 &=& < 1, 1 > \\ \\ \mathbf{v}_1 – \mathbf{v}_2 &=& < 5, -1 > - < -4, 2 > \\ \mathbf{v}_1 – \mathbf{v}_2 &=& < 5+4,-1 -2 > \\ \mathbf{v}_1 – \mathbf{v}_2 &=& < 9, -3 > \\ \\ \mathbf{v}_2 – \mathbf{v}_1 &=& < -4, 2 >- < 5, -1 > \\ \mathbf{v}_2 – \mathbf{v}_1 &=& < -4-5, 2+1 > \\ \mathbf{v}_2 – \mathbf{v}_1 &=& < -9, 3 > \\ \hline \end{array} \)

Compute
\(\cos \left( \dfrac{\pi}{12} \right) + \sin\left( \dfrac{\pi}{12} \right)\).

Formula:

\(\cos \left( x \right) + \sin \left( x \right) =\sqrt{2}\sin \left(x+\dfrac{\pi}{4} \right)\)

\(\begin{array}{|rcll|} \hline \mathbf{\cos \left( \dfrac{\pi}{12} \right) + \sin \left( \dfrac{\pi}{12} \right)} &=& \sqrt{2}\sin \left(\dfrac{\pi}{12}+\dfrac{\pi}{4} \right) \\ &=& \sqrt{2}\sin \left(\dfrac{4\pi}{12} \right) \\ &=& \sqrt{2}\sin \left(\dfrac{\pi}{3} \right) \quad | \quad \sin \left(\dfrac{\pi}{3} \right) = \dfrac{\sqrt{3}}{2} \\ &=& \sqrt{2} \dfrac{\sqrt{3}}{2} \\ &=& \mathbf{\dfrac{\sqrt{6}}{2} } \\ \hline \end{array} \)

Compute

\( \sin \dfrac{\pi}{12} + \sin \dfrac{3\pi}{12} + \sin \dfrac{5\pi}{12} + \sin \dfrac{7\pi}{12} + \sin \dfrac{9\pi}{12} + \sin \dfrac{11\pi}{12}\).

Formula: \(\sin(x)+\sin(y) = 2 \sin \left( \dfrac{x+y}{2} \right) \cos \left( \dfrac{x-y}{2} \right) \\ \cos(x)+\cos(y) = 2 \cos \left( \dfrac{x+y}{2} \right) \cos \left( \dfrac{x-y}{2} \right)\)

\(\small{ \begin{array}{|rcll|} \hline && \mathbf{\sin \left( \dfrac{\pi}{12} \right) + \sin \left( \dfrac{3\pi}{12} \right) + \sin \left( \dfrac{5\pi}{12} \right) + \sin \left( \dfrac{7\pi}{12} \right) + \sin \left( \dfrac{9\pi}{12} \right) + \sin \left( \dfrac{11\pi}{12} \right)} \\\\ &=& \Bigg( \sin \left( \dfrac{11\pi}{12} \right) + \sin \left( \dfrac{\pi}{12} \right) \Bigg) + \Bigg( \sin \left( \dfrac{9\pi}{12} \right) + \sin \left( \dfrac{3\pi}{12} \right)\Bigg) + \Bigg( \sin \left( \dfrac{7\pi}{12} \right) + \sin \left( \dfrac{5\pi}{12} \right) \Bigg)\\\\ &=& 2 \sin \left( \dfrac{ \dfrac{11\pi}{12}+\dfrac{\pi}{12}}{2} \right) \cos \left( \dfrac{ \dfrac{11\pi}{12}-\dfrac{\pi}{12}}{2} \right) +2 \sin \left( \dfrac{ \dfrac{9\pi}{12}+\dfrac{\pi}{3}}{2} \right) \cos \left( \dfrac{ \dfrac{11\pi}{9}-\dfrac{\pi}{3}}{2} \right) +2 \sin \left( \dfrac{ \dfrac{7\pi}{12}+\dfrac{\pi}{5}}{2} \right) \cos \left( \dfrac{ \dfrac{7\pi}{12}-\dfrac{\pi}{5}}{2} \right) \\\\ &=& 2 \sin \left( \dfrac{\pi}{2} \right) \cos \left( \dfrac{5\pi}{12} \right) +2 \sin \left( \dfrac{\pi}{2} \right) \cos \left( \dfrac{3\pi}{12} \right) +2 \sin \left( \dfrac{\pi}{2} \right) \cos \left( \dfrac{\pi}{12} \right) \\\\ &=& 2 \sin \left( \dfrac{\pi}{2} \right) \left( \cos \left( \dfrac{5\pi}{12} \right) +\cos \left( \dfrac{3\pi}{12} \right)+ \cos \left( \dfrac{\pi}{12} \right)\right) \quad | \quad \sin \left( \dfrac{\pi}{2} \right) = 1 \\\\ &=& 2 \left( \cos \left( \dfrac{5\pi}{12} \right) +\cos \left( \dfrac{3\pi}{12} \right)+ \cos \left( \dfrac{\pi}{12} \right)\right) \\\\ &=& 2 \left( \cos \left( \dfrac{5\pi}{12} \right) +\cos \left( \dfrac{\pi}{12} \right) + \cos \left( \dfrac{3\pi}{12} \right)\right) \\\\ &=& 2 \left( 2 \cos \left( \dfrac{ \dfrac{5\pi}{12}+\dfrac{\pi}{12}}{2} \right) \cos \left( \dfrac{ \dfrac{5\pi}{12}-\dfrac{\pi}{12}}{2} \right) + \cos \left( \dfrac{3\pi}{12} \right)\right) \\\\ &=& 2 \left( 2 \cos \left( \dfrac{3\pi}{12} \right) \cos \left( \dfrac{\pi}{6} \right) + \cos \left( \dfrac{3\pi}{12} \right)\right) \\\\ &=& 2 \cos \left( \dfrac{3\pi}{12} \right) \left( 2 \cos \left( \dfrac{\pi}{6} \right) + 1 \right) \\\\ &=& 2 \cos \left( \dfrac{\pi}{4} \right) \left( 2 \cos \left( \dfrac{\pi}{6} \right) + 1 \right) \quad | \quad \cos \left( \dfrac{\pi}{4} \right) = \dfrac{\sqrt{2}}{2},\ \cos \left( \dfrac{\pi}{6} \right) = \dfrac{\sqrt{3}}{2} \\\\ &=& 2* \dfrac{\sqrt{2}}{2} \left( 2* \dfrac{\sqrt{3}}{2} + 1 \right) \\\\ &=& \sqrt{2}(\sqrt{3} + 1 ) \\\\ &=& \mathbf{\sqrt{2}+\sqrt{6}} \\ \hline \end{array} }\)

1.

Calculate the unit vector for v = <3, 4>

\(\mathbf{v_{\text{unit}}} = < \dfrac{3} { \sqrt{3^2+4^2} } ,\ \dfrac{4}{\sqrt{3^2+4^2}} > \\ \mathbf{v_{\text{unit}}} = < \dfrac{3} {5} ,\ \dfrac{4}{5} >\)

2.

Find a vector with length 10 in the same direction of v.

\(\begin{array}{|rcll|} \hline && 10\mathbf{v_{\text{unit}}} \\ &=& 10*< \dfrac{3} {5} ,\ \dfrac{4}{5} > \\ &=& < \dfrac{10*3} {5} ,\ \dfrac{10*4}{5} > \\ &=& < \dfrac{30} {5} ,\ \dfrac{40}{5} > \\ &=& \mathbf{ < 6 ,\ 8 > } \\ \hline \end{array}\)

Simplify

\(\left(1 - \dfrac{1}{4} \right) \left(1 - \dfrac{1}{9} \right) \left(1 - \dfrac{1}{16} \right) \left(1 - \dfrac{1}{25} \right) \left(1 - \dfrac{1}{36} \right) \times \cdots \times \left(1 - \dfrac{1}{9801}\right) \left(1 - \dfrac{1}{10000}\right)\).

\(\small{ \begin{array}{|rcll|} \hline && \mathbf{\left(1 - \dfrac{1}{4} \right) \left(1 - \dfrac{1}{9} \right) \left(1 - \dfrac{1}{16} \right) \left(1 - \dfrac{1}{25} \right) \left(1 - \dfrac{1}{36} \right)\cdots \left(1 - \dfrac{1}{9801}\right) \left(1 - \dfrac{1}{10000}\right)} \\ \\ &=& \left(\dfrac{4-1}{4} \right) \left(\dfrac{9-1}{9} \right) \left(\dfrac{16-1}{16} \right) \left(\dfrac{25-1}{25} \right) \left(\dfrac{36-1}{36} \right)\cdots \left(\dfrac{9801-1}{9801}\right) \left(\dfrac{10000-1}{10000}\right) \\ \\ &=& \dfrac{(2-1)(2+1)}{2*2} \cdot \dfrac{(3-1)(3+1)}{3*3} \cdot \dfrac{(4-1)(4+1)}{4*4} \cdot \dfrac{(5-1)(5+1)}{5*5} \cdot \dfrac{(6-1)(6+1)}{6*6}\cdots \dfrac{(99-1)(99+1)}{99*99}\cdot \dfrac{(100-1)(100+1)}{100*100} \\ \\ &=& \dfrac{ 1*3 }{2*2} \cdot \dfrac{ 2*4 }{3*3} \cdot \dfrac{ 3*5 }{4*4} \cdot \dfrac{ 4*6 }{5*5} \cdot \dfrac{ 5*7 }{6*6}\cdots \dfrac{ 98*100 }{99*99}\cdot \dfrac{ 99*101 }{100*100} \\ \\ &=& \dfrac{ 1*\color{red}3 }{2*2} \cdot \dfrac{ {\color{green}2}*4 }{3*3} \cdot \dfrac{ 3*5 }{4*4} \cdot \dfrac{ 4*6 }{5*5} \cdot \dfrac{ 5*7 }{6*6}\cdots \dfrac{ 98*100 }{99*99}\cdot \dfrac{ 99*101 }{100*100} \quad | \quad \text{change red and green} \\ \\ &=& \dfrac{ 1*\color{green}2 }{2*2} \cdot \dfrac{ {\color{red}3}*4 }{3*3} \cdot \dfrac{ 3*5 }{4*4} \cdot \dfrac{ 4*6 }{5*5} \cdot \dfrac{ 5*7 }{6*6}\cdots \dfrac{ 98*100 }{99*99}\cdot \dfrac{ 99*101 }{100*100} \\ \\ &=& \dfrac{ 1*2}{2*2} \cdot \dfrac{ 3*{\color{red}4} }{3*3} \cdot \dfrac{ {\color{green}3} *5 }{4*4} \cdot \dfrac{ 4*6 }{5*5} \cdot \dfrac{ 5*7 }{6*6}\cdots \dfrac{ 98*100 }{99*99}\cdot \dfrac{ 99*101 }{100*100} \quad | \quad \text{change red and green} \\ \\ &=& \dfrac{ 1*2}{2*2} \cdot \dfrac{ 3*{\color{green}3} }{3*3} \cdot \dfrac{ {\color{red}4} *5 }{4*4} \cdot \dfrac{ 4*6 }{5*5} \cdot \dfrac{ 5*7 }{6*6}\cdots \dfrac{ 98*100 }{99*99}\cdot \dfrac{ 99*101 }{100*100} \\ \\ &=& \dfrac{ 1*2 }{2*2} \cdot \dfrac{ 3*3 }{3*3} \cdot \dfrac{ 4*{\color{red}5} }{4*4} \cdot \dfrac{ {\color{green}4} *6 }{5*5} \cdot \dfrac{ 5*7 }{6*6}\cdots \dfrac{ 98*100 }{99*99}\cdot \dfrac{ 99*101 }{100*100} \quad | \quad \text{change red and green} \\ \\ &=& \dfrac{ 1*2 }{2*2} \cdot \dfrac{ 3*3 }{3*3} \cdot \dfrac{ 4*{\color{green}4} }{4*4} \cdot \dfrac{ {\color{red}5} *6 }{5*5} \cdot \dfrac{ 5*7 }{6*6}\cdots \dfrac{ 98*100 }{99*99}\cdot \dfrac{ 99*101 }{100*100} \\ \\ &=& \dfrac{ 1*2 }{2*2} \cdot \dfrac{ 3*3 }{3*3} \cdot \dfrac{ 4*4 }{4*4} \cdot \dfrac{ 5*{\color{red}6} }{5*5} \cdot \dfrac{ {\color{green}5}*7 }{6*6}\cdots \dfrac{ 98*100 }{99*99}\cdot \dfrac{ 99*101 }{100*100} \quad | \quad \text{change red and green} \\ \\ &=& \dfrac{ 1*2 }{2*2} \cdot \dfrac{ 3*3 }{3*3} \cdot \dfrac{ 4*4 }{4*4} \cdot \dfrac{ 5*{\color{green}5} }{5*5} \cdot \dfrac{ {\color{red}6}*7 }{6*6}\cdots \dfrac{ 98*100 }{99*99}\cdot \dfrac{ 99*101 }{100*100} \\ \\ &=& \dfrac{ 1*2 }{2*2} \cdot \dfrac{ 3*3 }{3*3} \cdot \dfrac{ 4*4 }{4*4} \cdot \dfrac{ 5*5 }{5*5} \cdot \dfrac{ 6*7 }{6*6}\cdots \dfrac{ 98*100 }{99*99}\cdot \dfrac{ 99*101 }{100*100} \\ \\ && \ldots \\ \\ &=& \dfrac{ 1*2 }{2*2} \cdot \dfrac{ 3*3 }{3*3} \cdot \dfrac{ 4*4 }{4*4} \cdot \dfrac{ 5*5 }{5*5} \cdot \dfrac{ 6*6 }{6*6}\cdots \dfrac{ 99*{\color{red}100} }{99*99}\cdot \dfrac{ {\color{green}99}*101 }{100*100} \quad | \quad \text{change red and green} \\ \\ && \ldots \\ &=& \dfrac{ 1*2 }{2*2} \cdot \dfrac{ 3*3 }{3*3} \cdot \dfrac{ 4*4 }{4*4} \cdot \dfrac{ 5*5 }{5*5} \cdot \dfrac{ 6*6 }{6*6}\cdots \dfrac{ 99*{\color{green}99} }{99*99}\cdot \dfrac{ {\color{red}100}*101 }{100*100} \\ \\ &=& \dfrac{ 1*2 }{2*2} \cdot \dfrac{ 3*3 }{3*3} \cdot \dfrac{ 4*4 }{4*4} \cdot \dfrac{ 5*5 }{5*5} \cdot \dfrac{ 6*6 }{6*6}\cdots \dfrac{ 99*99 }{99*99}\cdot \dfrac{ 100*101 }{100*100} \\ \\ &=& \dfrac{1}{2}\cdot \dfrac{101}{100} \\\\ &=& \mathbf{\dfrac{101}{200}} \\ \hline \end{array} }\)

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

5.

Find cos A

cos-rule

\(\begin{array}{|rcll|} \hline 2^2 &=& 3^2+4^2-2*3*4*\cos(A) \\\\ 4 &=& 9+16-24\cos(A) \\ 4 &=& 25-24\cos(A) \\ 24\cos(A) &=& 25- 4 \\ 24\cos(A) &=& 21 \\ \cos(A) &=& \dfrac{21}{24} \\ \mathbf{\cos(A)} &=& \mathbf{\dfrac{3}{8} } \\ \hline \end{array}\)

4.

Find sin B

\(\begin{array}{|rcll|} \hline \cos(C) &=& \dfrac{6}{10} \\\\ \mathbf{\cos{C}} &=& \mathbf{\dfrac{3}{5}} \\ \hline \sin(C) &=& \sqrt{1-\cos^2(C)} \\ \sin(C) &=& \sqrt{1-\dfrac{9}{25}} \\ \sin(C) &=& \sqrt{\dfrac{16}{25}} \\ \mathbf{\sin(C)} &=& \mathbf{ \dfrac{4}{5} } \\ \hline \end{array}\)

cos-rule

\(\begin{array}{|rcll|} \hline AB^2 &=& 10^2 + (15+6)^2 -2*10(15+6)\cos(C) \quad | \quad \cos(C)=\dfrac{3}{5} \\\\ AB^2 &=& 100 + 441 -2*10*21*\dfrac{3}{5} \\ AB^2 &=& 541 -2*2*21*3 \\ AB^2 &=& 541 -252 \\ AB^2 &=& 289 \\ \mathbf{AB} &=& \mathbf{17} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \dfrac{\sin(B)}{10} &=& \dfrac{\sin(C)}{AB} \\\\ \sin(B) &=& \dfrac{10\sin(C)}{AB} \quad | \quad \sin(C)=\dfrac{4}{5},\ AB = 17 \\\\ \sin(B) &=& \dfrac{10*\dfrac{4}{5}}{17} \\\\ \sin(B) &=& \dfrac{2*4}{17} \\\\ \mathbf{ \sin(B) } &=& \mathbf{ \dfrac{8}{17} } \\ \hline \end{array}\)