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There is a row of Pascal's triangle that has three successive positive entries,"a" "b"  and "c" such that "b" is double "c"  and "a" is triple "c" If this row begins "1,n,"  then find n.

Guest May 21, 2017
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 #1
avatar+18777 
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There is a row of Pascal's triangle that has three successive positive entries,"a" "b"  and "c"

such that "b" is double "c" 

and "a" is triple "c"

If this row begins "1,n,"  then find n.

 

Three successive positive entries:

\(\begin{array}{rcll} a&=&\binom{n}{k-1} \\ b&=&\binom{n}{k} \\ c&=&\binom{n}{k+1} \\ \end{array} \)

 

"b" is double "c" and "a" is triple "c"

\(\begin{array}{|rcll|} \hline a &= 3c &=& \binom{n}{k-1} \\ b &= 2c &=& \binom{n}{k} \\ c & &=& \binom{n}{k+1} \\ \hline \end{array} \)

 

\(\begin{array}{|lrcll|} \hline (1) & 2c &=& \binom{n}{k} \quad & | \quad c = \binom{n}{k+1} \\ & 2\cdot \binom{n}{k+1} &=& \binom{n}{k} \quad & | \quad \binom{n}{k+1}= ( \frac{n-k}{k+1} ) \binom{n}{k} \\ & 2\cdot ( \frac{n-k}{k+1} ) \binom{n}{k} &=& \binom{n}{k} \\ & 2\cdot ( \frac{n-k}{k+1} ) &=& 1 \\ & \mathbf{ n-k } & \mathbf{=} & \mathbf{ \frac{k+1}{2} } \\\\ (2) & 3c &=& \binom{n}{k-1} \quad & | \quad c = \binom{n}{k+1} \\ & 3\cdot \binom{n}{k+1} &=& \binom{n}{k-1} \quad & | \quad \binom{n}{k+1}= ( \frac{n-k}{k+1} ) \binom{n}{k} \\ & 3\cdot ( \frac{n-k}{k+1} ) \binom{n}{k} &=& \binom{n}{k-1} \quad & | \quad \binom{n}{k-1}= ( \frac{k}{n-k+1} ) \binom{n}{k} \\ & 3\cdot ( \frac{n-k}{k+1} ) \binom{n}{k} &=& ( \frac{k}{n-k+1} ) \binom{n}{k} \\ & 3\cdot ( \frac{n-k}{k+1} ) &=& \frac{k}{n-k+1} \\ & 3\cdot (n-k)\cdot (n-k+1) &=& k\cdot (k+1) \quad & | \quad \mathbf{ n-k } \mathbf{=} \mathbf{ \frac{k+1}{2} } \\ & 3\cdot ( \frac{k+1}{2} )\cdot ( \frac{k+1}{2} +1) &=& k\cdot (k+1) \\ & 3\cdot ( \frac{k+1}{2} )\cdot ( \frac{k+3}{2} ) &=& k\cdot (k+1) \\ & \frac34\cdot (k+1)\cdot (k+3) &=& k\cdot (k+1) \\ & \frac34 \cdot (k+3) &=& k \\ & \frac34 k + \frac94 &=& k \\ & k-\frac34 k &=& \frac94 \\ & \frac14 k &=& \frac94 \\ & \mathbf{ k } & \mathbf{=} & \mathbf{9} \\\\ & \mathbf{ n-k } & \mathbf{=} & \mathbf{ \frac{k+1}{2} } \\ & n-9 & = & \frac{9+1}{2} \\ & n-9 & = & 5 \\ & \mathbf{ n } & \mathbf{=} & \mathbf{ 14 } \\ \hline \end{array}\)

 

The three successive positive entries are:

\(a=3003 =\binom{14}{8} \\ b=2002 =\binom{14}{9} \\ c=1001 =\binom{14}{10} \\ \)

and n is 14.

 

laugh

heureka  May 22, 2017
edited by heureka  May 22, 2017
 #2
avatar+79894 
0

Thanks, heureka....I am very intersted in this problem....your answer includes some combinatorial identities........as I am not too familiar with these, could you show me how these two are derived ??

 

[ n / ( k + 1) ]  =  [  (n - k ) / ( k + 1) ]   *  [ n / k ]

 

And

 

[ n / (k - 1) ]  =  [ k / (n - k + 1) ] * [ n / k ]

 

 

Also......what allows us to do this ??

 

2 [ (n - k) / (k + 1) ]  =  1

 

n - k   =  [ k + 1 ] / 2

 

Thanks....!!!!

 

 

cool cool cool

CPhill  Oct 25, 2017

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