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Question: The coefficient of the \({{\left( r-1 \right)}^{th}},{{r}^{th}}\) and \({{\left( r+1 \right)}^{th}}\...

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

Explanation

Solution

We know the (r+1)th{{\left( r+1 \right)}^{th}} term in expansion of (1+x)n{{\left( 1+x \right)}^{n}} is given by nCr(1)r(x)n−r^{n}{{C}_{r}}{{\left( 1 \right)}^{r}}{{\left( x \right)}^{n-r}} . Now we will hence find (r−1)th,rth{{\left( r-1 \right)}^{th}},{{r}^{th}} and (r+1)th{{\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 (1+x)n{{\left( 1+x \right)}^{n}} .
We know by binomial theorem the expansion of (a+b)n{{\left( a+b \right)}^{n}} the coefficient of (r+1)th{{\left( r+1 \right)}^{th}} term is given by nCr(a)r(b)n−r^{n}{{C}_{r}}{{\left( a \right)}^{r}}{{\left( b \right)}^{n-r}}
Hence we can say (r+1)th{{\left( r+1 \right)}^{th}} term in expansion of (1+x)n{{\left( 1+x \right)}^{n}} is given by nCr(1)r(x)n−r^{n}{{C}_{r}}{{\left( 1 \right)}^{r}}{{\left( x \right)}^{n-r}}
(r−1)th,rth{{\left( r-1 \right)}^{th}},{{r}^{th}} and (r+1)th{{\left( r+1 \right)}^{th}} in the expansion of (1+x)n{{\left( 1+x \right)}^{n}} are in ratio 1 : 3 : 5.
Hence we can say the ratio of (r−1)th{{\left( r-1 \right)}^{th}} and rth{{r}^{th}} term is 1 : 3.
Now we know that (r−1)th{{\left( r-1 \right)}^{th}} term in the expansion of (1+x)n{{\left( 1+x \right)}^{n}} is nCr−2^{n}{{C}_{r-2}} similarly in the expansion of rth{{r}^{th}} is given by nCr−1^{n}{{C}_{r-1}}
Hence we get nCr−2nCr−1=13\dfrac{^{n}{{C}_{r-2}}}{^{n}{{C}_{r-1}}}=\dfrac{1}{3} .
Now we know that nCr=n!(n−r)!r!^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!} hence using this we get
n!(n−(r−2))!(r−2)!n!(n−(r−1))!(r−1)!=13 ⇒1(n−r+2)!(r−2)!1(n−r+1)!(r−1)!=13 \begin{aligned} & \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{aligned}
⇒(n−r+1)!(r−1)!(n−r+2)!(r−2)!=13\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(n−1)!n!=n\left( n-1 \right)!

& \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{aligned}$$ Cross multiplying the equation we get $\begin{aligned} & 3r-3=n-r+2 \\\ & \Rightarrow 0=n-r+2-3r+3 \\\ & \Rightarrow n-4r+5=0 \\\ & \therefore n-4r=-5............................\left( 1 \right) \\\ \end{aligned}$ Now again consider the given condition ${{\left( r-1 \right)}^{th}},{{r}^{th}}$ and ${{\left( r+1 \right)}^{th}}$ in the expansion of $${{\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{aligned} & \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{aligned}$$ Now using $n!=n\left( n-1 \right)!$ we get $$\begin{aligned} & \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{aligned}$$ On cross multiplication we get $\begin{aligned} & \Rightarrow 5r=3n-3r+3 \\\ & \Rightarrow 0=3n-3r-5r+3 \\\ & \therefore 3n-8r=-3....................\left( 2 \right) \\\ \end{aligned}$ Multiplying equation (1) by 3 and then subtracting it from equation (2) we get $\begin{aligned} & 3n-8r-\left( 3n-12r \right)=-3-\left( -15 \right) \\\ & \Rightarrow 3n-3n-8r+12r=15-3 \\\ & \Rightarrow 4r=12 \\\ \end{aligned}$ Dividing the whole equation by 4 we get r = 3. Now if we substitute r = 3 in equation (1) we get $\begin{aligned} & n-4\left( 3 \right)=-5 \\\ & \Rightarrow n-12=-5 \\\ & \Rightarrow n=12-5 \\\ & \therefore n=7 \\\ \end{aligned}$ **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.