heureka

Compute the sum:
\(101^2-97^2+93^2-89^2+\ldots+5^2-1^2\)

\(\begin{array}{|rcll|} \hline && \mathbf{ 101^2-97^2+93^2-89^2+\ldots+5^2-1^2} \\\\ &=& 101^2 + 93^2 + 85^2 +\ldots+ 13^2+5^2 \\ && -97^2 -89^2 -81^2 -\ldots- 9^2-1^2 \\\\ &=& \mathbf{\sum \limits_{k=1}^{13}\Big(~ 8k-3~\Big)^2 -\sum \limits_{k=1}^{13}\Big(~ 8k-7~\Big)^2} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\sum \limits_{k=1}^{13}\Big(~ 8k-3~\Big)^2 -\sum \limits_{k=1}^{13}\Big(~ 8k-7~\Big)^2} \\\\ &=& \sum \limits_{k=1}^{13}\Big(~ 8k-3~\Big)^2 - \Big(~ 8k-7~\Big)^2 \\\\ &=& \sum \limits_{k=1}^{13}64k^2-48k+9-(64k^2-112k+49) \\\\ &=& \sum \limits_{k=1}^{13} {\color{red}64k^2}-48k+9{\color{red}-64k^2}+112k-49 \\\\ &=& \sum \limits_{k=1}^{13} -48k+9+112k-49 \\\\ &=& \sum \limits_{k=1}^{13} 64k-40 \\\\ &=& 64\sum \limits_{k=1}^{13}k -\sum \limits_{k=1}^{13}40 \quad | \quad \sum \limits_{k=1}^{13}40 = 40*13 \\\\ &=& 64\sum \limits_{k=1}^{13}k - 40*13 \quad | \quad \sum \limits_{k=1}^{13}k = \dfrac{(1+13)}{2}*13 \\\\ &=& 64\dfrac{(1+13)}{2}*13 - 40*13 \\\\ &=& 32*14*13 - 40*13 \\\\ &=& 5824 - 520 \\\\ &=& \mathbf{5304} \\ \hline \end{array}\)

\(\mathbf{ 101^2-97^2+93^2-89^2+\ldots+5^2-1^2} = \mathbf{5304}\)

Points \(A\), \(B\), and \(C\) are on sides \(\overline{WX}\), \(\overline{YZ}\), and \(\overline{XY}\) of rectangle \(WXYZ\) as shown below such that
\(XA = 4\), \(YB = 18\),  \(\angle ACB= 90^\circ\), and \(YC = 2XC\).
Find \(AB\).

\(\begin{array}{|rcll|} \hline \tan(A) = \dfrac{4}{XC} &=& \dfrac{2XC}{18} \\\\ \dfrac{4}{XC} &=& \dfrac{2XC}{18} \\\\ \dfrac{4}{XC} &=& \dfrac{XC}{9} \\\\ 36 &=& XC^2 \\\\ 6 &=& XC \\ \mathbf{XC} &=& \mathbf{6} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline x^2 &=& (3XC)^2 + BD^2 \quad | \quad BD =YB-XA \\ x^2 &=& (3XC)^2 + (YB-XA)^2 \\ x^2 &=& (3*6)^2 + (18-4)^2 \\ x^2 &=& 18^2 + 14^2 \\ x^2 &=& 324 + 196 \\ x^2 &=& 520 \\ x^2 &=& 4*130 \\ x &=& 2\sqrt{130} \\ \mathbf{x} &=& \mathbf{22.8035085020} \\ \hline \end{array} \)

\(\mathbf{AB \approx 22.8}\)

Find

\(99 - 97 + 95 - 93 + \ldots+ 3 - 1\).

\(\begin{array}{|l|rc|} \hline & \text{pair} \\ \hline 1 & 3-1 & = 2 \\ 2 & 7-5 & = 2 \\ 3 & 11-9 & = 2 \\ 4 & 15-13 & = 2 \\ 5 & 19-17 & = 2 \\ 6 & 23-21 & = 2 \\ 7 & 27-25 & = 2 \\ 8 & 31-29 & = 2 \\ 9 & 35-33 & = 2 \\ 10 & 39-37 & = 2 \\ 11 & 43-41 & = 2 \\ 12 & 47-45 & = 2 \\ 13 & 51-49 & = 2 \\ 14 & 55-53 & = 2 \\ 15 & 59-57 & = 2 \\ 16 & 63-61 & = 2 \\ 17 & 67-65 & = 2 \\ 18 & 71-69 & = 2 \\ 19 & 75-73 & = 2 \\ 20 & 79-77 & = 2 \\ 21 & 83-81 & = 2 \\ 22 & 87-85 & = 2 \\ 23 & 91-89 & = 2 \\ 24 & 95-93 & = 2 \\ 25 & 99-97 & = 2 \\ \hline \end{array}\)

\(99 - 97 + 95 - 93 + \ldots+ 3 - 1 = \mathbf{25 \times 2 = 50}\)

Jill is making a magic multiplication square using the numbers 1, 2, 4, 5, 10, 20, 25, 50, and 100.

The products of the numbers in each row, in each column, and in the two diagonals must all be the same.

In the figure you can see how she has started. Which number must go in the cell with the question mark?

Compute the sum:
\((a +(2n+1)d)^2- (a + (2n)d)^2 +(a + (2n-1)d)^2 - (a+(2n-2)d)^2 + \cdots + (a+d)^2 - a^2\).

\(\small{ \begin{array}{|rcll|} \hline && \mathbf{\Big(a +(2n+1)d\Big)^2- \Big(a + (2n)d\Big)^2 +\Big(a + (2n-1)d\Big)^2 - \Big(a+(2n-2)d\Big)^2 + \cdots + (a+d)^2 - a^2} \\\\ &=& \Big(a +(2n+1)d\Big)^2 +\Big(a + (2n-1)d\Big)^2+\Big(a + (2n-3)d\Big)^2 + \cdots + \Big(a+(2n-(2n-1))d\Big)^2 \\ && -\Big(a + (2n-0)d\Big)^2- \Big(a+(2n-2)d\Big)^2- \Big(a+(2n-4)d\Big)^2 - \cdots -\Big(a+(2n-(2n))d\Big)^2 \\\\ &=& \mathbf{\sum \limits_{k=1}^{n+1}\Big(~ a+(2k-1)d~\Big)^2 -\sum \limits_{k=1}^{n+1}\Big(~ a+2(k-1)d~\Big)^2} \\ \hline \end{array} }\)

\(\begin{array}{|rcll|} \hline && \mathbf{\sum \limits_{k=1}^{n+1}\Big(~ a+(2k-1)d~\Big)^2 -\sum \limits_{k=1}^{n+1}\Big(~ a+2(k-1)d~\Big)^2} \\\\ &=& \sum \limits_{k=1}^{n+1}\Big(~ a+(2k-1)d~\Big)^2 - \Big(~a+2(k-1)d~\Big)^2 \\\\ &=& \sum \limits_{k=1}^{n+1} a^2+2ad(2k-1)+d^2(2k-1)^2-\Big(~ a^2+4ad(k-1)+4d^2(k-1)^2 ~\Big) \\\\ &=& \sum \limits_{k=1}^{n+1} {\color{red}{a^2}}+2ad(2k-1)+d^2(2k-1)^2 {\color{red}{-a^2}}-4ad(k-1)-4d^2(k-1)^2 \\\\ &=& \sum \limits_{k=1}^{n+1} 2ad(2k-1)+d^2(2k-1)^2 -4ad(k-1)-4d^2(k-1)^2 \\\\ &=& \sum \limits_{k=1}^{n+1} {\color{red}{4adk}}-2ad+d^2(2k-1)^2 {\color{red}{-4adk}}+4ad-4d^2(k-1)^2 \\\\ &=& \sum \limits_{k=1}^{n+1} -2ad+d^2(2k-1)^2 +4ad-4d^2(k-1)^2 \\\\ &=& \sum \limits_{k=1}^{n+1} 2ad+d^2(2k-1)^2 -4d^2(k-1)^2 \\\\ &=& \sum \limits_{k=1}^{n+1}2ad +d^2~\sum \limits_{k=1}^{n+1} (2k-1)^2 -4(k-1)^2 \quad | \quad \sum \limits_{k=1}^{n+1}2ad = 2ad(n+1) \\\\ &=& 2ad(n+1) +d^2\sum \limits_{k=1}^{n+1} (2k-1)^2 -4(k-1)^2 \\\\ &=& 2ad(n+1) +d^2\sum \limits_{k=1}^{n+1} 4k^2-4k+1-4(k^2-2k+1) \\\\ &=& 2ad(n+1) +d^2\sum \limits_{k=1}^{n+1} {\color{red}{4k^2}}-4k+1{\color{red}{-4k^2}}+8k-4 \\\\ &=& 2ad(n+1) +d^2\sum \limits_{k=1}^{n+1} 4k-3 \\\\ &=& 2ad(n+1) +d^2\Big(~4\sum \limits_{k=1}^{n+1}k-\sum \limits_{k=1}^{n+1}3~\Big)\quad | \quad \sum \limits_{k=1}^{n+1}3 = 3(n+1) \\\\ &=& 2ad(n+1) +d^2\Big(~4\sum \limits_{k=1}^{n+1}k-3(n+1)~\Big) \\\\ &=& 2ad(n+1) +d^2\Big(~4\sum \limits_{k=1}^{n+1}k-3(n+1)~\Big) \quad | \quad \sum \limits_{k=1}^{n+1}k = \dfrac{1+(n+1)}{2}(n+1) \\\\ &=& 2ad(n+1) +d^2\Big(~4\dfrac{(n+2))}{2}(n+1)-3(n+1)~\Big) \\\\ &=& 2ad(n+1) +d^2\Big(~2(n+2)(n+1)-3(n+1)~\Big) \\\\ &=& 2ad(n+1) +d^2(n+1)\Big(~2(n+2)-3~\Big) \\\\ &=& 2ad(n+1) +d^2(n+1)(2n+4-3) \\\\ &=& \mathbf{2ad(n+1) +d^2(n+1)(2n+1)} \\ \hline \end{array}\)

\(\mathbf{\Big(a +(2n+1)d\Big)^2- \Big(a + (2n)d\Big)^2 +\Big(a + (2n-1)d\Big)^2 - \Big(a+(2n-2)d\Big)^2 + \cdots + (a+d)^2 - a^2} \\= \mathbf{2ad(n+1) +d^2(n+1)(2n+1)}\)

Compute
\(\dfrac{1^2}{2^1} + \dfrac{2^2}{2^2} + \dfrac{3^2}{2^3} +~ \ldots~ + \dfrac{n^2}{2^n} +~ \ldots\)

\(\small{ \begin{array}{|rcllll|} \hline \text{Sum}&=& \mathbf{\dfrac{1^2}{2^1}} &\mathbf{+\dfrac{2^2}{2^2}} &\mathbf{+ \dfrac{3^2}{2^3}} &\mathbf{+ \dfrac{4^2}{2^4}} &\mathbf{+~ \ldots~ + \dfrac{n^2}{2^n} +~ \ldots } \\\\ &=& 1^2\left(\dfrac{1}{2}\right) &+2^2\left(\dfrac{1}{2}\right)^2 &+3^2\left(\dfrac{1}{2}\right)^3 &+4^2\left(\dfrac{1}{2}\right)^4 &+~ \ldots~ + n^2\left(\dfrac{1}{2}\right)^n \quad | \quad r = \dfrac{1}{2} \\\\ &=& 1^2r&+2^2r^2&+3^2r^3&+4^2r^4 &+~ \ldots~ + n^2r^n +~ \ldots~ \\\\ &=& 1r &+4r^2 &+9r^3 &+16r^4 &+~ \ldots~ \\\\ &=& 1r &+1r^2 &+1r^3 &+1r^4 &+~ \ldots~ \quad | \quad s_1=r(1+r+r^2+r^3+~ \ldots~ ) \\ & & &+3r^2 &+3r^3 &+3r^4 &+~ \ldots~ \quad | \quad s_2=3r^2(1+r+r^2+r^3+~ \ldots~ ) \\ & & & &+5r^3 &+5r^4 &+~ \ldots~ \quad | \quad s_3=5r^3(1+r+r^2+r^3+~ \ldots~ ) \\ & & & & &+7r^4 &+~ \ldots~ \quad | \quad s_4=7r^4(1+r+r^2+r^3+~ \ldots~ ) \\ & & & & &\ldots & +~ \ldots~ \quad | \quad \ldots \\ \hline \end{array} \\ \begin{array}{|rcllll|} \hline &=& \left(s_1+s_2+s_3+s_4+~ \ldots~\right) \\ &=& \left(r+3r^2+5r^3+7r^4+~ \ldots~\right) \left(1+r+r^2+r^3+~ \ldots~ \right) \quad | \quad 1+r+r^2+r^3+~ \ldots = \dfrac{1}{1-r} \\ &=& \dfrac{1}{1-r}\left(r+3r^2+5r^3+7r^4+~ \ldots~\right) \\ \mathbf{\text{Sum}} &=& \mathbf{\dfrac{1}{1-r}S } \quad | \quad S= r+3r^2+5r^3+7r^4+~ \ldots \\ \hline \end{array} }\)

\(\small{ \begin{array}{|rcllll|} \hline \mathbf{S}&=& \mathbf{r} &\mathbf{+3r^2} &\mathbf{+5r^3} &\mathbf{+7r^4} &+~ \ldots \\ &=& r &+r^2 &+r^3 &+r^4 &+~ \ldots \quad | \quad S_1=r(1+r+r^2+r^3+~ \ldots~ ) \\ & & &+2r^2 &+2r^3 &+2r^4 &+~ \ldots \quad | \quad S_2=2r^2(1+r+r^2+r^3+~ \ldots~ ) \\ & & & &+2r^3 &+2r^4 &+~ \ldots \quad | \quad S_3=2r^3(1+r+r^2+r^3+~ \ldots~ ) \\ & & & & &+2r^4 &+~ \ldots \quad | \quad S_4=2r^4(1+r+r^2+r^3+~ \ldots~ ) \\ & & & & &\ldots& +~ \ldots~ \quad | \quad \ldots \\ \hline \end{array} \\ \begin{array}{|rcllll|} \hline &=& \left(S_1+S_2+S_3+S_4+~ \ldots~\right) \\ &=& \left(r+2r^2+2r^3+2r^4+~ \ldots~\right) \left(1+r+r^2+r^3+~ \ldots~ \right) \quad | \quad 1+r+r^2+r^3+~ \ldots = \dfrac{1}{1-r} \\ &=& \dfrac{1}{1-r}\left(r+2r^2+2r^3+2r^4+~ \ldots~\right) \\ &=& \dfrac{r}{1-r}\left(1+2r+2r^2+2r^3+~ \ldots~\right) \\ &=& \dfrac{r}{1-r}\left(1+2r(1+r+r^2+r^3~ \ldots~)\right) \quad | \quad 1+r+r^2+r^3+~ \ldots = \dfrac{1}{1-r} \\ &=& \dfrac{r}{1-r}\left(1+2r\left(\dfrac{1}{1-r}\right)\right) \\ &=& \dfrac{r}{1-r}\left(1+\dfrac{2r}{1-r}\right) \\ \mathbf{S}&=& \mathbf{\dfrac{r}{1-r}\left(\dfrac{1+r}{1-r}\right)} \\ \hline \end{array} }\)

\(\begin{array}{|rcll|} \hline \mathbf{\text{Sum}} &=& \mathbf{\left(\dfrac{1}{1-r}\right)S } \quad | \quad \mathbf{S=\dfrac{r}{1-r}\left(\dfrac{1+r}{1-r}\right)} \\\\ \text{Sum}&=&\left(\dfrac{1}{1-r}\right)\left(\dfrac{r}{1-r}\right)\left(\dfrac{1+r}{1-r}\right) \\\\ \text{Sum}&=&\dfrac{r(1+r)}{(1-r)^3} \quad | \quad r=\dfrac{1}{2} \\\\ \text{Sum}&=&\dfrac{\dfrac{1}{2}\left(1+\dfrac{1}{2}\right)}{\left(1-\dfrac{1}{2}\right)^3} \\\\ \text{Sum}&=&\dfrac{\dfrac{1}{2}\left(\dfrac{3}{2}\right)}{\left(\dfrac{1}{2}\right)^3} \\\\ \text{Sum}&=&\dfrac{\left(\dfrac{3}{2}\right)}{\left(\dfrac{1}{2}\right)^2} \\\\ \text{Sum}&=&\dfrac{\left(\dfrac{3}{2}\right)}{\left(\dfrac{1}{4}\right)} \\\\ \text{Sum}&=&\dfrac{12}{2} \\\\ \mathbf{\text{Sum}}&=& \mathbf{6} \\ \hline \end{array}\)

Geometry - Analytic Geometry

cosθ=−√2/3 , where π≤θ≤3π/2 .

tanβ=4/3 , where 0≤β≤π/2 .

What is the exact value of sin(θ+β) ?

Give your answer as a fraction in simplified form.

Consider the vectors \(\mathbf{v} = \begin{pmatrix} 1\\2\\1 \end{pmatrix}\), \(\mathbf{w} = \begin{pmatrix} 1\\4 \\5 \end{pmatrix}\), and  \(\mathbf{x} = \begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix}\).
If the vectors \(\mathbf{v}\), \(\mathbf{w}\) and \(\mathbf{x}\) are linearly independent, answer with ? (a question mark.)
If they aren't, find coefficients a,b and c, not all 0, such that \(a\begin{pmatrix} 1\\2\\1 \end{pmatrix}+b \begin{pmatrix} 1\\4 \\5 \end{pmatrix} + c\begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\)

and answer with \(\dfrac{a-b}{c}\).

\(\begin{array}{|rcll|} \hline \begin{array}{|rcll|} \hline a\begin{pmatrix} 1\\2\\1 \end{pmatrix}+b \begin{pmatrix} 1\\4 \\5 \end{pmatrix} + c\begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix} &=& \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \\ \hline \end{array}\\\\ \left(\begin{array}{@{}ccc|c@{}} 1 & 1 & -1& 0 \\ 2 & 4 & 6 & 0\\ 1 & 5 & 15 & 0 \end{array}\right)_{\frac{R_2}{2}\rightarrow R_2} \Rightarrow \left(\begin{array}{@{}ccc|c@{}} 1 & 1 & -1& 0 \\ 1 & 2 & 3 & 0\\ 1 & 5 & 15 & 0 \end{array}\right)_{R_3-R_1 \rightarrow R_3} \Rightarrow \\ \Rightarrow \left(\begin{array}{@{}ccc|c@{}} 1 & 1 & -1& 0 \\ 1 & 2 & 3 & 0\\ 0 & 4 & 16 & 0 \end{array}\right)_{\frac{R_3}{4}\rightarrow R_3} \Rightarrow \left(\begin{array}{@{}ccc|c@{}} 1 & 1 & -1& 0 \\ 1 & 2 & 3 & 0\\ 0 & 1 & 4 & 0 \end{array}\right)_{R_2-R_1 \rightarrow R_2} \Rightarrow \\ \Rightarrow \left(\begin{array}{@{}ccc|c@{}} 1 & 1 & -1& 0 \\ 0 & 1 & 4 & 0\\ 0 & 1 & 4 & 0 \end{array}\right)_{R_3-R_2\rightarrow R_3} \Rightarrow \left(\begin{array}{@{}ccc|c@{}} 1 & 1 & -1& 0 \\ 0 & 1 & 4 & 0\\ 0 & 0 & 0 & 0 \end{array}\right)_{R_1-R_2 \rightarrow R_1} \Rightarrow \\ \Rightarrow \left(\begin{array}{@{}ccc|c@{}} 1 & 0 & -5& 0 \\ 0 & 1 & 4 & 0\\ 0 & 0 & 0 & 0 \end{array}\right) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \left(\begin{array}{@{}ccc|c@{}} 1 & 0 & -5& 0 \\ 0 & 1 & 4 & 0\\ 0 & 0 & 0 & 0 \end{array}\right) \\ & \begin{array}{rcll} a -5c &=& 0 \\ b+4c &=& 0 \\ c &=& k \\ \end{array}\quad \Rightarrow \quad \begin{array}{rcll} a -5k &=& 0 \\ b+4k &=& 0 \\ c &=& k \\ \end{array}\quad \Rightarrow \quad \begin{array}{rcll} \mathbf{a} &=& \mathbf{5k} \\ \mathbf{b} &=& \mathbf{-4k} \\ \mathbf{c} &=& \mathbf{k} \\ \end{array} \\ \hline \end{array} \)

\((a,\ b,\ c) = (5k,\ -4k,\ k) \Rightarrow \text{Nontrivial}\Rightarrow \mathbf{\text{Linearly dependent}}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{a-b}{c}} &=& \dfrac{5k-(-4k)}{k} \\\\ \dfrac{a-b}{c} &=& \dfrac{9k}{k} \\\\ \mathbf{\dfrac{a-b}{c}} &=& \mathbf{9} \\ \hline \end{array}\)

If we let \(f(n)\) denote the sum of all the positive divisors of the integer \(n\),
how many integers i exist such that \(1 \le i \le 2010\) and \(f(i) = 1 + \sqrt{i} + i\)?

\(\small{ \begin{array}{|r|r|r|r|l|} \hline n & \text{prime number} & i=\text{prime number}^2 & \text{the sum of the positive divisors i } & f(i) = 1+\sqrt{i} + i \\ \hline 1 & 2 & 4 & 7 & f(2^2) = 1+2+2^2 = 7 \\ 2 & 3 & 9 & 13 & f(3^2) = 1+3+3^2 = 13 \\ 3 & 5 & 25 & 31 & f(5^2) = 1+5+5^2 = 31 \\ 4 & 7 & 49 & 57 & f(7^2) = 1+7+7^2 = 57 \\ 5 & 11 & 121 & 133 & f(11^2) = 1+11+11^2 = 133 \\ 6 & 13 & 169 & 183 & f(13^2) = 1+13+13^2 = 183 \\ 7 & 17 & 289 & 307 & f(17^2) = 1+17+17^2 = 307 \\ 8 & 19 & 361 & 381 & f(19^2) = 1+19+19^2 = 381 \\ 9 & 23 & 529 & 553 & f(23^2) = 1+23+23^2 = 553 \\ 10 & 29 & 841 & 871 & f(29^2) = 1+29+29^2 = 871 \\ 11 & 31 & 961 & 993 & f(31^2) = 1+31+31^2 = 993 \\ 12 & 37 &1396 & 1407 & f(37^2) = 1+37+37^2 = 1407 \\ 13 & 41 &1681 & 1723 & f(41^2) = 1+41+41^2 = 1723 \\ 14 & 43 &1849 & 1893 & f(43^2) = 1+43+43^2 = 1893 \\ \hline \end{array} }\)