Question

The coefficients of three consecutive terms of \[{\left( {1 + x} \right)^{n + 5}}\] are in the ratio \[5:10:14\], then \[n = \]

Answer 1

Hint: In this question, we will proceed by letting the consecutive terms and finding their coefficients. Then use the given ratio two find two equations in terms of \[n\]. Solve them to find the required answer.

 Complete step by step answer:
Let the three consecutive terms be \[{\left( {r - 1} \right)^{th}},{r^{th}}\] and \[{\left( {r + 1} \right)^{th}}\] terms i.e., \[{T_{r - 1}},{T_r},{T_{r + 1}}\] of the expansion \[{\left( {1 + x} \right)^{n + 5}}\].
We know that the general term of expansion \[{\left( {1 + x} \right)^n}\] is \[{T_{r + 1}} = {}^n{C_r}{x^r}\]
So, for the expansion \[{\left( {1 + x} \right)^{n + 5}}\] we have
\[
   \Rightarrow {T_{r - 1}} = {}^{n + 5}{C_{r - 2}}{x^{r - 2}} \\
   \Rightarrow {T_r} = {}^{n + 5}{C_{r - 1}}{x^{r - 1}} \\
   \Rightarrow {T_{r + 1}} = {}^{n + 5}{C_r}{x^r} \\
\]
Hence the coefficients of terms \[{\left( {r - 1} \right)^{th}},{r^{th}}\] and \[{\left( {r + 1} \right)^{th}}\] are \[{}^{n + 5}{C_{r - 2}},{}^{n + 5}{C_{r - 1}}\] and \[{}^{n + 5}{C_r}\] respectively.
Given that the coefficients of \[{\left( {r - 1} \right)^{th}},{r^{th}}\] and \[{\left( {r + 1} \right)^{th}}\] terms are in a ration of \[5:10:14\].
So, we have
\[
   \Rightarrow \dfrac{{{\text{coefficient of }}{{\left( {r - 1} \right)}^{th}}}}{{{\text{coefficient of }}{r^{th}}}} = \dfrac{5}{{10}} \\
   \Rightarrow \dfrac{{{}^{n + 5}{C_{r - 2}}}}{{{}^{n + 5}{C_{r - 1}}}} = \dfrac{5}{{10}} \\
\]
Using the formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] we have
\[
   \Rightarrow \dfrac{{\dfrac{{\left( {n + 5} \right)!}}{{\left( {r - 2} \right)!\left[ {n - \left( {r - 2} \right)} \right]!}}}}{{\dfrac{{\left( {n + 5} \right)!}}{{\left( {r - 1} \right)!\left[ {n - \left( {r - 1} \right)} \right]!}}}} = \dfrac{5}{{10}} \\
   \Rightarrow \dfrac{{\left( {r - 1} \right)!\left[ {n - \left( {r - 1} \right)} \right]!}}{{\left( {r - 2} \right)!\left[ {n - \left( {r - 2} \right)} \right]!}} = \dfrac{5}{{10}} \\
   \Rightarrow \dfrac{{\left( {r - 1} \right)\left( {r - 2} \right)!\left[ {n - \left( {r - 1} \right)} \right]!}}{{\left( {r - 2} \right)!\left( {n - \left( {r - 2} \right)} \right)\left[ {n - \left( {r - 1} \right)} \right]!}} = \dfrac{5}{{10}} \\
   \Rightarrow \dfrac{{\left( {r - 1} \right)}}{{n - \left( {r - 2} \right)}} = \dfrac{5}{{10}} \\
   \Rightarrow 10r - 10 = 5n - 5r + 10 \\
   \Rightarrow 5n = 15r - 20....................................................\left( 1 \right) \\
\]
Also, we have
\[
   \Rightarrow \dfrac{{{\text{coefficient of }}{r^{th}}}}{{{\text{coefficient of }}{{\left( {r + 1} \right)}^{th}}}} = \dfrac{{10}}{{14}} \\
   \Rightarrow \dfrac{{{}^{n + 5}{C_{r - 2}}}}{{{}^{n + 5}{C_{r - 1}}}} = \dfrac{5}{7} \\
\]
Using the formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] we have
\[
   \Rightarrow \dfrac{{\dfrac{{\left( {n + 5} \right)!}}{{\left( {r - 1} \right)!\left[ {n - \left( {r - 1} \right)} \right]!}}}}{{\dfrac{{\left( {n + 5} \right)!}}{{\left( r \right)!\left[ {n - \left( r \right)} \right]!}}}} = \dfrac{5}{7} \\
   \Rightarrow \dfrac{{\left( r \right)!\left[ {n - \left( r \right)} \right]!}}{{\left( {r - 1} \right)!\left[ {n - \left( {r - 1} \right)} \right]!}} = \dfrac{5}{7} \\
   \Rightarrow \dfrac{{\left( r \right)\left( {r - 1} \right)!\left[ {n - \left( r \right)} \right]!}}{{\left( {r - 1} \right)!\left( {n - \left( {r - 1} \right)} \right)\left[ {n - \left( r \right)} \right]!}} = \dfrac{5}{7} \\
   \Rightarrow \dfrac{{\left( r \right)}}{{n - \left( {r - 1} \right)}} = \dfrac{5}{7} \\
   \Rightarrow 7r = 5n - 5r + 5 \\
   \Rightarrow 5n - 12r + 5 = 0 \\
   \Rightarrow 5n = 12r - 5....................................................\left( 2 \right) \\
\]
From equation (1) and (2), we have
\[
   \Rightarrow 15r - 20 = 12r - 5 \\
   \Rightarrow 15r - 12r = 20 - 5 \\
   \Rightarrow 3r = 15 \\
   \Rightarrow r = \dfrac{{15}}{3} \\
  \therefore r = 5 \\
\]
Substituting \[r = 5\] in equation (1), we have
\[
   \Rightarrow 5n = 15\left( 5 \right) - 20 \\
   \Rightarrow 5n = 75 - 20 \\
   \Rightarrow 5n = 55 \\
   \Rightarrow n = \dfrac{{55}}{5} \\
  \therefore n = 11 \\
\]
Thus, the value of \[n\] is 11.

Note:
The general term of expansion \[{\left( {1 + x} \right)^n}\] is \[{T_{r + 1}} = {}^n{C_r}{x^r}\]. Don’t confuse when you are expanding the terms of combinations. We can also check our answer by substituting the values of \[r,n\] in the considered terms by checking whether their coefficients are consecutive or not.