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# binomial question

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

(b) For some positive integer $$n$$ the expansion of $$(1 + x)^n$$ has three consecutive coefficients $$a,b,c$$ that satisfy $$a:b:c = 1:7:35.$$ What must $$n$$ be?

Jul 20, 2021

#1
+26226
+2

(a) Simplify

$$\dfrac{\dbinom{n}{k}}{\dbinom{n}{k - 1}}$$.

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

Jul 21, 2021
#3
0

Thank you so much for answering my question but could you please elaborate on how you got $$(n-k+1)!=(n-k)!(n-k+1)$$

Guest Jul 28, 2021
#2
+26226
+1

(b)

For some positive integer $$n$$ the expansion of $$(1 + x)^n$$
has three consecutive coefficients $$a,~b,~c$$ that satisfy
$$a:b:c = 1:7:35$$.
What must $$n$$ be?

$$\text{Let a=\dbinom{n}{k - 1} } \\ \text{Let b=\dbinom{n}{k} } \\ \text{Let c=\dbinom{n}{k+1} }$$

$$\begin{array}{|rcll|} \hline \dfrac{b}{a} = \dfrac{7}{1} &=& \dfrac{\dbinom{n}{k}}{\dbinom{n}{k - 1}} \\\\ \dfrac{7}{1} &=& \dfrac{\dbinom{n}{k}}{\dbinom{n}{k - 1}} \\\\ \dfrac{7}{1} &=& \dfrac{n-k+1}{k} \\\\ 7k &=& n-k+1 \\ 8k &=& n+1 \\ \mathbf{n} &=& \mathbf{8k-1} \\ \hline \end{array}$$

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

$$\begin{array}{|rcll|} \hline \dfrac{c}{b} = \dfrac{35}{7} &=& \dfrac{\dbinom{n}{k+1}}{\dbinom{n}{k}} \\\\ \dfrac{35}{7} &=& \dfrac{\dbinom{n}{k} *\dfrac{(n-k)}{(k+1)}}{\dbinom{n}{k}} \\\\ \dfrac{35}{7} &=&\dfrac{(n-k)}{(k+1)} \\\\ 35(k+1) &=&7(n-k) \\ 42k &=& 7n-35 \\ \mathbf{k} &=& \mathbf{ \dfrac{7n-35}{42}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline n &=& 8k-1 \quad | \quad \mathbf{k=\dfrac{7n-35}{42}} \\\\ n &=& 8\left(\dfrac{7n-35}{42}\right)-1 \\\\ n &=& \dfrac{4(7n-35)}{21}-1 \\\\ n &=& \dfrac{4(7n-35)-21}{21} \\\\ 21n &=&4(7n-35)-21\\ 21n &=&28n-140-21\\ 7n &=& 161 \quad | \quad :7 \\ \mathbf{n} &=& \mathbf{23} \\ \hline k &=& \dfrac{7n-35}{42} \\\\ k &=& \dfrac{7*23-35}{42} \\\\ \mathbf{k} &=& \mathbf{3} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline a:b:c =\dbinom{n}{k - 1}:\dbinom{n}{k}:\dbinom{n}{k + 1} &=& 1:7:35 \\\\ \dbinom{n}{k - 1}:\dbinom{n}{k}:\dbinom{n}{k + 1} &=& 1:7:35 \\\\ \dbinom{23}{3 - 1}:\dbinom{23}{3}:\dbinom{23}{3 + 1} &=& 1:7:35 \\\\ \dbinom{23}{2}:\dbinom{23}{3}:\dbinom{23}{4} &=& 1:7:35 \\\\ 253:1771:8855 &=& 1:7:35 \\ \hline \end{array}$$

Jul 21, 2021