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Question: If the ration of the \({{7}^{th}}\) term from the beginning to the \({{7}^{th}}\) term from the end ...

If the ration of the 7th{{7}^{th}} term from the beginning to the 7th{{7}^{th}} term from the end in the expansion of (213+1313)n{{\left( {{2}^{\dfrac{1}{3}}}+\dfrac{1}{{{3}^{\dfrac{1}{3}}}} \right)}^{n}} is 16\dfrac{1}{6} then, find the value of n.

Explanation

Solution

To solve this problem we need to know about the binomial expansion of the term (213+1313)n{{\left( {{2}^{\dfrac{1}{3}}}+\dfrac{1}{{{3}^{\dfrac{1}{3}}}} \right)}^{n}}. In binomial (r+1)th{{\left( r+1 \right)}^{th}} term from beginning of the expression (a+b)n{{\left( a+b \right)}^{n}} is given by nCran−rbr^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}. Now if we reverse the expression i.e. as (b+a)n{{\left( b+a \right)}^{n}} then the (r+1)th{{\left( r+1 \right)}^{th}} term nCr−1bn−rar^{n}{{C}_{r-1}}{{b}^{n-r}}{{a}^{r}} will be equal to (r+1)th{{\left( r+1 \right)}^{th}} term from the end of the expression (a+b)n{{\left( a+b \right)}^{n}}. Now we will apply the above mentioned formulas in our given question to solve it.

Complete step-by-step answer:
To solve this question we should know that the binomial expansion of the (a+b)n{{\left( a+b \right)}^{n}} is given by nCran−rbr^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}} where r represents its (r+1)th{{\left( r+1 \right)}^{th}} term.
Now we are given the expression as,
(213+1313)n{{\left( {{2}^{\dfrac{1}{3}}}+\dfrac{1}{{{3}^{\dfrac{1}{3}}}} \right)}^{n}}
Hence its (r+1)th{{\left( r+1 \right)}^{th}} term will be given by
=nCr(213)n−r(1313)r{{=}^{n}}{{C}_{r}}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{n-r}}{{\left( \dfrac{1}{{{3}^{\dfrac{1}{3}}}} \right)}^{r}}
So 7th{{7}^{th}} term from the beginning will be given when r = 6, so we get
=nC6(213)n−6(1313)6{{=}^{n}}{{C}_{6}}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{n-6}}{{\left( \dfrac{1}{{{3}^{\dfrac{1}{3}}}} \right)}^{6}}
=nC6(2)n−63(13)2{{=}^{n}}{{C}_{6}}{{\left( 2 \right)}^{\dfrac{n-6}{3}}}{{\left( \dfrac{1}{3} \right)}^{2}}
Now when we reverse the expression like (1313+213)n{{\left( \dfrac{1}{{{3}^{\dfrac{1}{3}}}}+{{2}^{\dfrac{1}{3}}} \right)}^{n}}. Now (r+1)th{{\left( r+1 \right)}^{th}} term from the beginning of this expression will give us the (r+1)th{{\left( r+1 \right)}^{th}} term from the end of (213+1313)n{{\left( {{2}^{\dfrac{1}{3}}}+\dfrac{1}{{{3}^{\dfrac{1}{3}}}} \right)}^{n}} ,
So 7th{{7}^{th}} term from the end of (213+1313)n{{\left( {{2}^{\dfrac{1}{3}}}+\dfrac{1}{{{3}^{\dfrac{1}{3}}}} \right)}^{n}}, we get as
=nCr(1313)n−r(213)r{{=}^{n}}{{C}_{r}}{{\left( \dfrac{1}{{{3}^{\dfrac{1}{3}}}} \right)}^{n-r}}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{r}}
Now putting r = 6, we get

& {{=}^{n}}{{C}_{6}}{{\left( \dfrac{1}{{{3}^{\dfrac{1}{3}}}} \right)}^{n-6}}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{6}} \\\ & {{=}^{n}}{{C}_{6}}{{\left( \dfrac{1}{3} \right)}^{\dfrac{n-6}{3}}}{{\left( 2 \right)}^{\dfrac{6}{3}}} \\\ \end{aligned}$$ Now it is given that the ratio of ${{7}^{th}}$ term from the beginning and ${{7}^{th}}$ term from the end is $\dfrac{1}{6}$, So we get, $\dfrac{^{n}{{C}_{6}}{{\left( 2 \right)}^{\dfrac{n-6}{3}}}{{\left( \dfrac{1}{3} \right)}^{2}}}{^{n}{{C}_{6}}{{\left( \dfrac{1}{3} \right)}^{\dfrac{n-6}{3}}}{{\left( 2 \right)}^{\dfrac{6}{3}}}}=\dfrac{1}{6}$ $6.\left( ^{n}{{C}_{6}}{{\left( 2 \right)}^{\dfrac{n-6}{3}}}{{\left( \dfrac{1}{3} \right)}^{2}} \right){{=}^{n}}{{C}_{6}}{{\left( \dfrac{1}{3} \right)}^{\dfrac{n-6}{3}}}{{\left( 2 \right)}^{2}}$ Cancelling out $^{n}{{C}_{6}}$ from both sides we get, $$\begin{aligned} & \Rightarrow 6.{{\left( 2 \right)}^{\dfrac{n-6}{3}}}{{\left( \dfrac{1}{3} \right)}^{2}}={{\left( \dfrac{1}{3} \right)}^{\dfrac{n-6}{3}}}{{\left( 2 \right)}^{2}} \\\ & \Rightarrow 6.{{\left( 2 \right)}^{\dfrac{n-6}{3}}}{{3}^{-2}}={{3}^{\dfrac{6-n}{3}}}{{.2}^{2}} \\\ \end{aligned}$$ Further solving we get, $$\begin{aligned} & \Rightarrow 6.{{\left( 2 \right)}^{\dfrac{n}{3}}}\times \dfrac{1}{36}=\dfrac{36}{{{3}^{\dfrac{n}{3}}}} \\\ & \Rightarrow {{6}^{\dfrac{n}{3}}}={{6}^{3}} \\\ \end{aligned}$$ Now coefficient of both sides is same so equating the powers we get, $\begin{aligned} & \dfrac{n}{3}=3 \\\ & n=9 \\\ \end{aligned}$ Hence we get our answer as 9. **Note:** To solve this problem you should be well aware of the binomial expansion theorem and you should know how to find the term from the end of the binomial expansion otherwise you would not be able to solve this so remember the mentioned formulas of binomial expansion in the above solution for future problems.