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(a) Simplify \(\frac{\binom{n}{k}}{\binom{n}{k - 1}}\)

(b) In the expansion of \((1 + x)^n,\)three consecutive coefficients are in the ratio 1:7:35. Find the positive integer n.

 Jan 8, 2020
 #1
avatar+24344 
+3

(a)

Simplify \(\dfrac{\dbinom{n}{k}}{\dbinom{n}{k - 1}}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\dbinom{n}{k - 1}} &=& \dfrac{n!}{(k-1)!(n-k+1)!} \quad &| \quad (k-1)!k=k!~ \text{or}~(k-1)!= \dfrac{k!}{k} \\\\ &=& k *\dfrac{n!}{k!(n-k+1)!} \quad &| \quad (n-k)!(n-k+1)=(n-k+1)! \\\\ &=& \dfrac{k}{n-k+1} * \dfrac{n!}{k!(n-k)!} \quad &| \quad \dfrac{n!}{k!(n-k)!} = \dbinom{n}{k} \\\\ &=& \mathbf{ \dfrac{k}{n-k+1} \dbinom{n}{k} } \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{\dbinom{n}{k}}{\dbinom{n}{k - 1}}} \\\\ &=& \dfrac{\dbinom{n}{k}}{\dfrac{k}{n-k+1} \dbinom{n}{k} } \\\\ &=& \dfrac{n-k+1}{k} * \dfrac{\dbinom{n}{k}}{\dbinom{n}{k} } \\\\ &=& \mathbf{\dfrac{n-k+1}{k}} \\ \hline \end{array}\)

 

 

laugh

 Jan 8, 2020
 #2
avatar+24344 
+3

(b)

In the expansion of \(\left(1 + x\right)^n\), three consecutive coefficients are in the ratio \(1:7:35\).
Find the positive integer \(n\).

 

\(\begin{array}{|rcll|} \hline \mathbf{\dbinom{n}{k + 1}} &=& \dfrac{n!}{(k+1)!(n-k-1)!} \quad &| \quad k!(k+1)=(k+1)! \\\\ &=& \dfrac{1}{k+1} *\dfrac{n!}{k!(n-k-1)!} \quad &| \quad (n-k-1)!(n-k)=(n-k)! \\\\ &=& \dfrac{n-k}{k+1} * \dfrac{n!}{k!(n-k)!} \quad &| \quad \dfrac{n!}{k!(n-k)!} = \dbinom{n}{k} \\\\ &=& \mathbf{ \dfrac{n-k}{k+1} \dbinom{n}{k} } \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \dbinom{n}{k - 1} : \dbinom{n}{k} : \dbinom{n}{k + 1} &=& 1 : 7 : 35 \\\\ &&\boxed{ \mathbf{\dbinom{n}{k - 1} = \dfrac{k}{n-k+1} \dbinom{n}{k} }\\ ~\\ \mathbf{\dbinom{n}{k + 1} = \dfrac{n-k}{k+1} \dbinom{n}{k} } } \\\\ \dfrac{k}{n-k+1} \dbinom{n}{k} : \dbinom{n}{k} : \dfrac{n-k}{k+1} \dbinom{n}{k} &=& 1 : 7 : 35 \\ \hline \dfrac{\dbinom{n}{k}}{\dfrac{k}{n-k+1} \dbinom{n}{k}} &=& \dfrac{7}{1} \\\\ \dfrac{n-k+1}{k} &=& 7 \\\\ n-k+1 &=& 7k \\ \mathbf{n} &=& \mathbf{8k -1} \qquad (1) \\ \hline \dfrac{\dfrac{n-k}{k+1} \dbinom{n}{k} } {\dbinom{n}{k}} &=& \dfrac{35}{7} \\\\ \dfrac{n-k}{k+1} &=& 5 \\\\ 5(k+1) &=& n-k \\ 5k+5 &=& n-k \\ 6k &=& n-5 \\ \mathbf{k} &=& \mathbf{ \dfrac{n-5}{6} } \qquad (2) \\ \hline n &=& 8k-1 \quad | \quad k=\dfrac{n-5}{6} \\ n &=& 8 \dfrac{(n-5)}{6}-1 \\ n &=& \dfrac{4}{3}(n-5)-1 \\ & & \dotsb \\ \mathbf{n} &=& \mathbf{23} \\\\ k &=& \dfrac{n-5}{6} \quad | \quad n=23 \\\\ k &=& \dfrac{23-5}{6} \\ & & \dotsb \\ \mathbf{k} &=& \mathbf{3} \\ \hline \dbinom{n}{k - 1} : \dbinom{n}{k} : \dbinom{n}{k + 1} &=& 1 : 7 : 35 \\\\ \dbinom{23}{2} : \dbinom{23}{3} : \dbinom{23}{4} &=& 1 : 7 : 35 \\\\ 253 : 1771 : 8855 &=& 1 : 7 : 35 \\ \hline \end{array}\)

 

laugh

 Jan 8, 2020
 #3
avatar+109061 
+1

Impressive, heureka   !!!!

 

 

cool cool cool

CPhill  Jan 8, 2020
 #4
avatar+24344 
+2

Thank you, CPhill !

 

laugh

heureka  Jan 9, 2020
 #5
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0

for the first panel, what does the line across the middle mean?

Guest Jan 10, 2020

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