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Hi, I don't understand how to do this question. Can someone please help and explain how to do it?

 

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

 Jul 29, 2023
 #1
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Let the 3 consecutive co-efficients be 

 

     \(\binom{n}{r-1},\qquad \binom{n}{r},\qquad \binom{n}{r+1}\\ \text{And these are in the ratio }\qquad 1:7:35\\~\\ so\\ \binom{n}{r-1}=\frac{1}{7}\cdot \binom{n}{r}\qquad \color{red}{(1)} \color{black}{\quad and } \qquad \binom{n}{r+1}=5\cdot \binom{n}{r}\quad \color{red}{(2)}\\~\\ \)

\(\binom{n}{r-1}=\frac{1}{7}\cdot \binom{n}{r}\qquad \color{red}{(1)} \\~\\ LHS=\frac{n!}{(r-1)!(n-r+1)!}\\~\\ LHS=\frac{n!}{\frac{r!}{(r)}(n-r)!(n-r+1)}\\~\\ LHS=\frac{n!\qquad (r)}{r!(n-r)!\quad (n-r+1)}\\~\\ LHS=\binom{n}{r}\frac{ (r)}{ (n-r+1)}\\~\\ so\\ \frac{1}{7}=\frac{r}{n-r+1}\\ n-r+1=7r\\ n=8r-1 \)

 

Now simplify equation 2 and then solve simulataneously to find n and r.     Then check your answer

 

 

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LaTex:

\binom{n}{r-1},\qquad \binom{n}{r},\qquad \binom{n}{r+1}\\
\text{And these are in the ratio  }\qquad 1:7:35\\~\\
so\\
\binom{n}{r-1}=\frac{1}{7}\cdot \binom{n}{r}\qquad \color{red}{(1)} \color{black}{\quad and } \qquad 
 \binom{n}{r+1}=5\cdot \binom{n}{r}\quad \color{red}{(2)}\\~\\

 

\binom{n}{r-1}=\frac{1}{7}\cdot \binom{n}{r}\qquad \color{red}{(1)} \\~\\
LHS=\frac{n!}{(r-1)!(n-r+1)!}\\~\\
LHS=\frac{n!}{\frac{r!}{(r)}(n-r)!(n-r+1)}\\~\\
LHS=\frac{n!\qquad (r)}{r!(n-r)!\quad (n-r+1)}\\~\\
LHS=\binom{n}{r}\frac{ (r)}{ (n-r+1)}\\~\\
so\\
\frac{1}{7}=\frac{r}{n-r+1}\\
n-r+1=7r\\
n=8r-1

 Jul 30, 2023

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