Question

In the binomial expression ${{\left( 1+\dfrac{k}{6} \right)}^{n}}$ what is the value of n given the fact that the coefficient of the third term is twice the coefficient of the fourth term?

Answer 1

Hint: Consider k as the variable and use the formula for the general term of a binomial expression ${{\left( a+b \right)}^{n}}$ given as ${{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$. Substitute the value of r = 2 and r = 3 to determine the third and fourth term. Equate their coefficients according to the information given in the question and solve for the value of n to get the answer. Reject the invalid values of n.

Complete step by step solution:
Here we have been provided with the binomial expression ${{\left( 1+\dfrac{k}{6} \right)}^{n}}$ and we are asked to find the value of n with the given condition that the coefficient of the third term is twice the coefficient of the fourth term.
Now, the general term of a binomial expression ${{\left( a+b \right)}^{n}}$ is given by the formula ${{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$, so substituting r = 2 in the formula we get,
$\begin{align}
  & \Rightarrow {{T}_{2+1}}={}^{n}{{C}_{2}}{{\left( 1 \right)}^{n-2}}{{\left( \dfrac{k}{6} \right)}^{2}} \\
 & \Rightarrow {{T}_{3}}=\dfrac{{}^{n}{{C}_{2}}}{{{6}^{2}}}{{\left( k \right)}^{2}} \\
\end{align}$
Substituting r = 3 in the formula of general term we get,
$\begin{align}
  & \Rightarrow {{T}_{3+1}}={}^{n}{{C}_{3}}{{\left( 1 \right)}^{n-3}}{{\left( \dfrac{k}{6} \right)}^{3}} \\
 & \Rightarrow {{T}_{4}}=\dfrac{{}^{n}{{C}_{3}}}{{{6}^{3}}}{{\left( k \right)}^{3}} \\
\end{align}$
According to the question we have the coefficient of the third term equal to twice the coefficient of the fourth term, so mathematically we have,
$\Rightarrow \dfrac{{}^{n}{{C}_{2}}}{{{6}^{2}}}=2\times \dfrac{{}^{n}{{C}_{3}}}{{{6}^{3}}}$
Using the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and cancelling the common factors we get,
$\Rightarrow \dfrac{n\left( n-1 \right)}{2!}=\dfrac{n\left( n-1 \right)\left( n-2 \right)}{3\times 3!}$
Now, we can cancel $n$ and $\left( n-1 \right)$ from both the sides because n cannot be equal to 0 and 1. If that happens then there will be only one and two terms respectively in the binomial expression but the information is given about the third and fourth terms. Therefore we get,
$\begin{align}
  & \Rightarrow \dfrac{1}{2}=\dfrac{\left( n-2 \right)}{3\times 6} \\
 & \Rightarrow 9=n-2 \\
 & \therefore n=11 \\
\end{align}$
Hence, the value of n is 11.

Note: Note that in the given expression k was the variable and we were equating its coefficients and not the overall term, that is why we did not consider the term containing the exponent of k. Always check each value if you get multiple answers of a variable as it may be possible that some of the values may be invalid that must be rejected.