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 #1
avatar+26400 
+12

Evaluate without a calculator:

\([~12\binom{3}{3} + 11\binom{4}{3} + 10\binom{5}{3} + \cdots + 2\binom{13}{3} +\binom{14}{3} ~] \)

 

Identity:

\(\text{For }~ n,r \in N^+, r \le n,\\ \dbinom{n+1}{r+1} = \sum \limits_{j=r}^{n} \dbinom{j}{r}\)

see: http://www.tau.ac.il/~tsirel/dump/Static/knowino.org/wiki/Pascal's_triangle.html

 

\(\small{ \begin{array}{lcl} \binom{3}{3} + \binom{4}{3} + \binom{5}{3}+ \binom{6}{3}+ \binom{7}{3}+ \binom{8}{3}+ \binom{9}{3}+ \binom{10}{3}+ \binom{11}{3}+ \binom{12}{3} + \binom{13}{3} +\binom{14}{3} &=& \binom{15}{4} \\ \binom{3}{3} + \binom{4}{3} + \binom{5}{3}+ \binom{6}{3}+ \binom{7}{3}+ \binom{8}{3}+ \binom{9}{3}+ \binom{10}{3}+ \binom{11}{3}+ \binom{12}{3} + \binom{13}{3} &=& \binom{14}{4} \\ \binom{3}{3} + \binom{4}{3} + \binom{5}{3}+ \binom{6}{3}+ \binom{7}{3}+ \binom{8}{3}+ \binom{9}{3}+ \binom{10}{3}+ \binom{11}{3}+ \binom{12}{3} &=& \binom{13}{4} \\ \binom{3}{3} + \binom{4}{3} + \binom{5}{3}+ \binom{6}{3}+ \binom{7}{3}+ \binom{8}{3}+ \binom{9}{3}+ \binom{10}{3}+ \binom{11}{3} &=& \binom{12}{4} \\ \binom{3}{3} + \binom{4}{3} + \binom{5}{3}+ \binom{6}{3}+ \binom{7}{3}+ \binom{8}{3}+ \binom{9}{3}+ \binom{10}{3} &=& \binom{11}{4} \\ \binom{3}{3} + \binom{4}{3} + \binom{5}{3}+ \binom{6}{3}+ \binom{7}{3}+ \binom{8}{3}+ \binom{9}{3} &=& \binom{10}{4} \\ \binom{3}{3} + \binom{4}{3} + \binom{5}{3}+ \binom{6}{3}+ \binom{7}{3}+ \binom{8}{3} &=& \binom{9}{4} \\ \binom{3}{3} + \binom{4}{3} + \binom{5}{3}+ \binom{6}{3}+ \binom{7}{3} &=& \binom{8}{4} \\ \binom{3}{3} + \binom{4}{3} + \binom{5}{3}+ \binom{6}{3} &=& \binom{7}{4} \\ \binom{3}{3} + \binom{4}{3} + \binom{5}{3} &=& \binom{6}{4} \\ \binom{3}{3} + \binom{4}{3} &=& \binom{5}{4} \\ \binom{3}{3} &=& \binom{4}{4} \end{array} } \)

 

\(\small{ \begin{array}{lcl} \binom{4}{4} +\binom{5}{4}+\binom{6}{4}+ \binom{7}{4}+ \binom{8}{4}+\binom{9}{4}+\binom{10}{4}+\binom{11}{4}+\binom{12}{4}+\binom{13}{4}+\binom{14}{4}+\binom{15}{4} \\ = \binom{16}{5}\\ 12\binom{3}{3} + 11\binom{4}{3} + 10\binom{5}{3}+9\binom{6}{3}+8\binom{7}{3}+7\binom{8}{3}+6\binom{9}{3}+5\binom{10}{3}+4\binom{11}{3}+3\binom{12}{3} + 2\binom{13}{3} +\binom{14}{3} \\ = \binom{16}{5} \\ 12\binom{3}{3} + 11\binom{4}{3} + 10\binom{5}{3}+9\binom{6}{3}+8\binom{7}{3}+7\binom{8}{3}+6\binom{9}{3}+5\binom{10}{3}+4\binom{11}{3}+3\binom{12}{3} + 2\binom{13}{3} +\binom{14}{3} \\ = \frac{16}{5}\cdot \frac{15}{4}\cdot \frac{14}{3}\cdot \frac{13}{2}\cdot \frac{12}{1} \\\\ \mathbf{ 12\binom{3}{3} + 11\binom{4}{3} + 10\binom{5}{3} +9\binom{6}{3}+8\binom{7}{3}+7\binom{8}{3}+6\binom{9}{3} +5\binom{10}{3}+4\binom{11}{3}+3\binom{12}{3} + 2\binom{13}{3} +\binom{14}{3} } \\ \mathbf{=} \mathbf{4368 } \end{array} } \)

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May 9, 2016
 #2
avatar+26400 
+5

if i have sides with each having a length of 13, 14, and 15. how big will each angle be?

 

 

Heron's formula states that the area of a triangle whose sides have lengths a, b, and c is
\(A = \sqrt{s(s-a)(s-b)(s-c)},\)
where s is the semiperimeter of the triangle; that is,
\(s=\frac{a+b+c}{2}\)

 

\(\begin{array}{rcl} s &=& \frac{a+b+c}{2} \\ &=& \frac{13+14+15}{2} \\ &=& \frac{42}{2} \\ \mathbf{s} & \mathbf{=}& \mathbf{21} \\\\ A &=& \sqrt{s\cdot (s-a)\cdot (s-b)\cdot (s-c)} \\ &=& \sqrt{21\cdot (21-13)\cdot (21-14)\cdot (21-15)} \\ &=& \sqrt{21\cdot 8 \cdot 7 \cdot 6} \\ &=& \sqrt{7056} \\ \mathbf{A} & \mathbf{=}& \mathbf{84} \end{array} \)

 

\(\begin{array}{rcl} 2\cdot A &=& b\cdot c\cdot \sin \alpha \\ \sin \alpha &=& \frac{2\cdot A}{ b\cdot c } \\ \sin \alpha &=& \frac{2\cdot 84}{ 14\cdot 15 } \\ \alpha &=& \arcsin{ (\frac{168}{ 14\cdot 15 } ) } \\ \alpha &=& \arcsin{ (0.8) } \\ \mathbf{\alpha }&\mathbf{=}& \mathbf{53.1301023542^{\circ}} \end{array} \)

 

\(\begin{array}{rcl} 2\cdot A &=& c\cdot a\cdot \sin \beta \\ \sin \beta &=& \frac{2\cdot A}{ c\cdot a } \\ \sin \beta &=& \frac{2\cdot 84}{ 15\cdot 13 } \\ \beta &=& \arcsin{ (\frac{168}{ 15\cdot 13 } ) }\\ \beta &=& \arcsin{ (0.86153846154) }\\ \mathbf{\beta} &\mathbf{=}& \mathbf{59.4897625939^{\circ}} \end{array} \)

 

\(\begin{array}{rcl} 2\cdot A &=& a\cdot b\cdot \sin \gamma \\ \sin \gamma &=& \frac{2\cdot A}{ a\cdot b} \\ \sin \gamma &=& \frac{2\cdot 84}{ 13\cdot 14 } \\ \gamma &=& \arcsin{ (\frac{168}{ 13\cdot 14 }) }\\ \gamma &=& \arcsin{ (0.92307692308) }\\ \mathbf{\gamma }&\mathbf{=}& \mathbf{67.3801350520^{\circ}} \end{array}\)

 

laugh

May 9, 2016
 #1
avatar+33665 
+5
May 9, 2016
 #1
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May 9, 2016

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