Question

The coefficient of the ${{\left( r-1 \right)}^{th}},{{r}^{th}}$ and ${{\left( r+1 \right)}^{th}}$ in the expansion of \[{{\left( 1+x \right)}^{n}}\] are in ratio 1 : 3 : 5. Find n and r.

Answer 1

Hint: We know the ${{\left( r+1 \right)}^{th}}$ term in expansion of \[{{\left( 1+x \right)}^{n}}\] is given by $^{n}{{C}_{r}}{{\left( 1 \right)}^{r}}{{\left( x \right)}^{n-r}}$ . Now we will hence find ${{\left( r-1 \right)}^{th}},{{r}^{th}}$ and ${{\left( r+1 \right)}^{th}}$ terms and take the ratios to form equations. Now we will solve the equation for n and r.

Complete step by step answer:
Now we are given the expression \[{{\left( 1+x \right)}^{n}}\] .
We know by binomial theorem the expansion of ${{\left( a+b \right)}^{n}}$ the coefficient of ${{\left( r+1 \right)}^{th}}$ term is given by $^{n}{{C}_{r}}{{\left( a \right)}^{r}}{{\left( b \right)}^{n-r}}$
Hence we can say ${{\left( r+1 \right)}^{th}}$ term in expansion of \[{{\left( 1+x \right)}^{n}}\] is given by $^{n}{{C}_{r}}{{\left( 1 \right)}^{r}}{{\left( x \right)}^{n-r}}$
${{\left( r-1 \right)}^{th}},{{r}^{th}}$ and ${{\left( r+1 \right)}^{th}}$ in the expansion of \[{{\left( 1+x \right)}^{n}}\] are in ratio 1 : 3 : 5.
Hence we can say the ratio of ${{\left( r-1 \right)}^{th}}$ and ${{r}^{th}}$ term is 1 : 3.
Now we know that ${{\left( r-1 \right)}^{th}}$ term in the expansion of \[{{\left( 1+x \right)}^{n}}\] is $^{n}{{C}_{r-2}}$ similarly in the expansion of ${{r}^{th}}$ is given by $^{n}{{C}_{r-1}}$
Hence we get $\dfrac{^{n}{{C}_{r-2}}}{^{n}{{C}_{r-1}}}=\dfrac{1}{3}$ .
Now we know that $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ hence using this we get
$\begin{align}
  & \dfrac{\dfrac{n!}{\left( n-\left( r-2 \right) \right)!\left( r-2 \right)!}}{\dfrac{n!}{\left( n-\left( r-1 \right) \right)!\left( r-1 \right)!}}=\dfrac{1}{3} \\
 & \Rightarrow \dfrac{\dfrac{1}{\left( n-r+2 \right)!\left( r-2 \right)!}}{\dfrac{1}{\left( n-r+1 \right)!\left( r-1 \right)!}}=\dfrac{1}{3} \\
\end{align}$
$\Rightarrow \dfrac{\left( n-r+1 \right)!\left( r-1 \right)!}{\left( n-r+2 \right)!\left( r-2 \right)!}=\dfrac{1}{3}$
Now we know $n!=n\left( n-1 \right)!$
\[\begin{align}
  & \Rightarrow \dfrac{\left( n-r+1 \right)!\left( r-1 \right)\left( r-2 \right)!}{\left( n-r+2 \right)\left( n-r+1 \right)!\left( r-2 \right)!}=\dfrac{1}{3} \\
 & \Rightarrow \dfrac{r-1}{n-r+2}=\dfrac{1}{3} \\
\end{align}\]
Cross multiplying the equation we get
$\begin{align}
  & 3r-3=n-r+2 \\
 & \Rightarrow 0=n-r+2-3r+3 \\
 & \Rightarrow n-4r+5=0 \\
 & \therefore n-4r=-5............................\left( 1 \right) \\
\end{align}$
Now again consider the given condition ${{\left( r-1 \right)}^{th}},{{r}^{th}}$ and ${{\left( r+1 \right)}^{th}}$ in the expansion of \[{{\left( 1+x \right)}^{n}}\] are in ratio 1 : 3 : 5.
Hence we have ${{r}^{th}}$ and ${{\left( r+1 \right)}^{th}}$ term is in the ratio 3 : 5
Hence we get $\dfrac{^{n}{{C}_{r-1}}}{^{n}{{C}_{r}}}=\dfrac{3}{5}$
Now again we will use the $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ , hence we have
$\dfrac{\dfrac{n!}{\left( n-\left( r-1 \right) \right)!\left( r-1 \right)!}}{\dfrac{n!}{\left( n-r \right)!r!}}=\dfrac{3}{5}$
\[\begin{align}
  & \Rightarrow \dfrac{\dfrac{1}{\left( n-r+1 \right)!\left( r-1 \right)!}}{\dfrac{1}{\left( n-r \right)!r!}}=\dfrac{3}{5} \\
 & \Rightarrow \dfrac{\left( n-r \right)!r!}{\left( n-r+1 \right)!\left( r-1 \right)!}=\dfrac{3}{5} \\
\end{align}\]
Now using $n!=n\left( n-1 \right)!$ we get
\[\begin{align}
  & \dfrac{\left( n-r \right)!r\left( r-1 \right)!}{\left( n-r+1 \right)\left( n-r \right)!\left( r-1 \right)!}=\dfrac{3}{5} \\
 & \Rightarrow \dfrac{r}{n-r+1}=\dfrac{3}{5} \\
\end{align}\]
On cross multiplication we get
$\begin{align}
  & \Rightarrow 5r=3n-3r+3 \\
 & \Rightarrow 0=3n-3r-5r+3 \\
 & \therefore 3n-8r=-3....................\left( 2 \right) \\
\end{align}$
Multiplying equation (1) by 3 and then subtracting it from equation (2) we get
$\begin{align}
  & 3n-8r-\left( 3n-12r \right)=-3-\left( -15 \right) \\
 & \Rightarrow 3n-3n-8r+12r=15-3 \\
 & \Rightarrow 4r=12 \\
\end{align}$
Dividing the whole equation by 4 we get r = 3.
Now if we substitute r = 3 in equation (1) we get
$\begin{align}
  & n-4\left( 3 \right)=-5 \\
 & \Rightarrow n-12=-5 \\
 & \Rightarrow n=12-5 \\
 & \therefore n=7 \\
\end{align}$

Hence we get n = 7 and r = 3.

Note: Now note that in the expansion of ${{\left( a+b \right)}^{n}}$ the coefficient of ${{\left( r+1 \right)}^{th}}$ term is given by $^{n}{{C}_{r}}{{\left( a \right)}^{r}}{{\left( b \right)}^{n-r}}$ hence to find ${{r}^{th}}$ term we will get $^{n}{{C}_{r-1}}{{\left( a \right)}^{r-1}}{{\left( b \right)}^{n-\left( r-1 \right)}}$ and not $^{n}{{C}_{r}}{{\left( a \right)}^{r}}{{\left( b \right)}^{n-r}}$ . Also note that for solving we can take the ratio of ${{\left( r-1 \right)}^{th}}$ and ${{\left( r+1 \right)}^{th}}$ terms also. Though it will be simple if we take consecutive terms as things cancel out and we get linear equations. In other cases we will have to deal with a 2 degree equation.