pascals triangle

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.

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....!!!!