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 #3
avatar+134 
0
Oct 28, 2019
 #3
avatar+26382 
0
Oct 28, 2019
 #6
avatar+26382 
+2

There exist positive integers n and k such that

\(32 \dbinom{6}{6} + 31 \dbinom{7}{6} + 30 \dbinom{8}{6} + \dots + 3 \dbinom{35}{6} + 2 \dbinom{36}{6} + \dbinom{37}{6} = \dbinom{n}{k}\)
Enter the ordered pair \((n, k)\)

 

Using the hockey stick identity: \(\sum \limits_{i=r}^{n} \dbinom{i}{r} = \dbinom{n+1}{r+1} \qquad \text{for } n,r\in \mathbb{N},\quad n \geq r\)

see: https://en.wikipedia.org/wiki/Hockey-stick_identity

 

\(\begin{array}{|rcll|} \hline &&\mathbf{ 32 \dbinom{6}{6} + 31 \dbinom{7}{6} + 30 \dbinom{8}{6} + \dots + 3 \dbinom{35}{6} + 2 \dbinom{36}{6} + \dbinom{37}{6} } \\\\ &=& \sum \limits_{i=6}^{37} \dbinom{i}{r} +\sum \limits_{i=6}^{36} \dbinom{i}{r} +\sum \limits_{i=6}^{35} \dbinom{i}{r} +\sum \limits_{i=6}^{34} \dbinom{i}{r} +\sum \limits_{i=6}^{33} \dbinom{i}{r} +\sum \limits_{i=6}^{32} \dbinom{i}{r} \\ && +\sum \limits_{i=6}^{31} \dbinom{i}{r} +\sum \limits_{i=6}^{30} \dbinom{i}{r} +\sum \limits_{i=6}^{29} \dbinom{i}{r} +\sum \limits_{i=6}^{28} \dbinom{i}{r} +\sum \limits_{i=6}^{27} \dbinom{i}{r} +\sum \limits_{i=6}^{26} \dbinom{i}{r} \\ && +\sum \limits_{i=6}^{25} \dbinom{i}{r} +\sum \limits_{i=6}^{24} \dbinom{i}{r} +\sum \limits_{i=6}^{23} \dbinom{i}{r} +\sum \limits_{i=6}^{22} \dbinom{i}{r} +\sum \limits_{i=6}^{21} \dbinom{i}{r} +\sum \limits_{i=6}^{20} \dbinom{i}{r} \\ && +\sum \limits_{i=6}^{19} \dbinom{i}{r} +\sum \limits_{i=6}^{18} \dbinom{i}{r} +\sum \limits_{i=6}^{17} \dbinom{i}{r} +\sum \limits_{i=6}^{16} \dbinom{i}{r} +\sum \limits_{i=6}^{15} \dbinom{i}{r} +\sum \limits_{i=6}^{14} \dbinom{i}{r} \\ && +\sum \limits_{i=6}^{13} \dbinom{i}{r} +\sum \limits_{i=6}^{12} \dbinom{i}{r} +\sum \limits_{i=6}^{11} \dbinom{i}{r} +\sum \limits_{i=6}^{10} \dbinom{i}{r} +\sum \limits_{i=6}^{9} \dbinom{i}{r} +\sum \limits_{i=6}^{8} \dbinom{i}{r} \\ && +\sum \limits_{i=6}^{7} \dbinom{i}{r} +\sum \limits_{i=6}^{6} \dbinom{i}{r} \\ \\ &=& \dbinom{38}{7} +\dbinom{37}{7} +\dbinom{36}{7} +\dbinom{35}{7} +\dbinom{34}{7} +\dbinom{33}{7} \\ && \dbinom{32}{7} +\dbinom{31}{7} +\dbinom{30}{7} +\dbinom{29}{7} +\dbinom{28}{7} +\dbinom{27}{7} \\ && \dbinom{26}{7} +\dbinom{25}{7} +\dbinom{24}{7} +\dbinom{23}{7} +\dbinom{22}{7} +\dbinom{21}{7} \\ && \dbinom{20}{7} +\dbinom{19}{7} +\dbinom{18}{7} +\dbinom{17}{7} +\dbinom{16}{7} +\dbinom{15}{7} \\ && \dbinom{14}{7} +\dbinom{13}{7} +\dbinom{12}{7} +\dbinom{11}{7} +\dbinom{10}{7} +\dbinom{9}{7} \\ && +\dbinom{8}{7} +\dbinom{7}{7} \\\\ &=& \sum \limits_{i=7}^{38} \dbinom{i}{r} \\\\ &=& \mathbf{\dbinom{39}{8}} \\ \hline \end{array}\)

 

laugh

Oct 28, 2019
 #1
avatar+129643 
+2

sin x + sin (3x)   =  -cos x  + cos (3x)

 

sinx  + sin2xcosx + sinxcos2x  = - cosx  + cos2xcosx - sin2xsin x

 

(sinx  + cosx )  =   - sin2x( six + cosx)  + cos2x (cosx - sin x)

 

(sinx + cosx)  = -2sinxcosx (sin x + cosx)  + (cosxcosx - sinxsinx)(cosx - sinx)

 

(sin x + cosx)  = -2sin^2xcosx - 2sinxcos^2x + cos^3x - sinxcos^2x  - sin^2xcosx + sin^3x

 

sin x + cosx  =  = -3sin^2xcosx - 3cos^2x sinx  +cos^3x + sin^3x

 

sin x + cosx  =  -3(1 - cos^2x)cosx  - 3(1- sin^2x)sinx  + cos^3x + sin^3x

 

sinx + cosx  = -3cosx+ 3cos^3x - 3sinx + 3sin^3x  + cos^3x + sin^3x

 

sin x + cos x  = -3(sinx + cos x)  + 4cos^3x + 4sin^3x

 

4 (sin x + cos x)  = 4 (sin^3x + cos^3x)       divide both sides by 4

 

sin x + cos x =  sin^3x + cos^3x            factor the right side as a sum of cubes

 

(sin x + cos x)  = (sin x + cos x) (sin^2x  - sinxcosx + cos^2x)

 

(sin x + cosx) =  (sin x + cosx) ( 1 - sin xcosx)

 

(sin x + cosx) (1 - sin xcosx - 1)  =  0

 

(sin x + cos x) ( -sin x cosx )  =  0

 

So

 

sin x + cosx  =  0                      and           -sinx cosx  = 0

x  =  3pi/4   or  7pi/4                                   sinx cos x   = 0                                                       

                                                              x  =  0   or   x   =  pi/2   or  x  = pi  or  x  = 3pi/2  or  x  = 2pi

 

 

 

cool cool cool   

Oct 28, 2019

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