# In the expansion of ${{\left( {{2}^{\dfrac{1}{3}}}+{{3}^{-\dfrac{1}{3}}} \right)}^{n}}$, the ratio of the ${{7}^{th}}$ term from the beginning to the ${{7}^{th}}$ term from the end is 1 : 6, then n is equal to –

(a) 6

(b) 7

(c) 8

(d) 9

Answer

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Hint: For solving this question, we need to know about the basic concepts of binomial theorem and the basic properties of algebraic expansion raised to the power of n. The formula for ${{(a+b)}^{n}}$ is given by $\sum\limits_{r=0}^{n}{\dfrac{n!}{r!\left( n-r \right)!}{{a}^{r}}{{b}^{n-r}}}$. We will use this to obtain the ${{7}^{th}}$ term from the beginning and the ${{7}^{th}}$ term from the end.

Complete step-by-step answer:

While solving the problem we should know the basics of binomial theorem and how to expand an algebraic expression raised to the power n. Now, we know that the formula for ${{(a+b)}^{n}}$ is given by $\sum\limits_{r=0}^{n}{\dfrac{n!}{r!\left( n-r \right)!}{{a}^{n-r}}{{b}^{r}}}$. Thus, in general, the ${{r}^{th}}$ from the beginning would be $\dfrac{n!}{(r-1)!\left( n-r+1 \right)!}{{a}^{n-r+1}}{{b}^{r-1}}$. Thus, to solve this problem, we have the ${{7}^{th}}$ term as –

=$\dfrac{n!}{6!\left( n-6 \right)!}{{a}^{n-6}}{{b}^{6}}$

(Now, we know that a =${{2}^{\dfrac{1}{3}}}$ and b = ${{3}^{-\dfrac{1}{3}}}$, since, the expression here is ${{\left( {{2}^{\dfrac{1}{3}}}+{{3}^{-\dfrac{1}{3}}} \right)}^{n}}$)

= \[\dfrac{n!}{6!\left( n-6 \right)!}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{n-6}}{{\left( {{3}^{-\dfrac{1}{3}}} \right)}^{6}}\] -- (1)

Now, the ${{7}^{th}}$ term from the end would be the ${{(n-6)}^{th}}$ term from the beginning. Thus, we have,

= $\dfrac{n!}{\left( n-6 \right)!6!}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{6}}{{\left( {{3}^{-\dfrac{1}{3}}} \right)}^{n-6}}$ -- (2)

Since, the ratio is 1:6. Thus, we have

$\dfrac{{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{n-6}}{{\left( {{3}^{-\dfrac{1}{3}}} \right)}^{6}}}{{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{6}}{{\left( {{3}^{-\dfrac{1}{3}}} \right)}^{n-6}}}=\dfrac{1}{6}$

Thus,

${{\left( {{2}^{\dfrac{1}{3}}} \right)}^{(n-6)-6}}{{\left( {{3}^{-\dfrac{1}{3}}} \right)}^{6-(n-6)}}=\dfrac{1}{6}$

${{\left( {{2}^{\dfrac{1}{3}}} \right)}^{n-12}}{{\left( {{3}^{-\dfrac{1}{3}}} \right)}^{12-n}}=\dfrac{1}{6}$

${{\left( {{2}^{-\dfrac{1}{3}}} \right)}^{12-n}}{{\left( {{3}^{-\dfrac{1}{3}}} \right)}^{12-n}}=\dfrac{1}{6}$

\[{{\left( {{2}^{-\dfrac{1}{3}}}\times {{3}^{-\dfrac{1}{3}}} \right)}^{12-n}}=\dfrac{1}{6}\]

\[{{\left( {{(2\times 3)}^{-\dfrac{1}{3}}} \right)}^{12-n}}=\dfrac{1}{6}\]

\[{{\left( {{6}^{-\dfrac{1}{3}}} \right)}^{12-n}}=\dfrac{1}{6}\]

\[{{\left( {{6}^{-\dfrac{1}{3}}} \right)}^{12-n}}={{6}^{-1}}\]

Now, we can compare the co-efficient to get the required value of n. We have,

$\dfrac{-(12-n)}{3}=-1$

-(12-n) = -3

12-n = 3

n = 9

Hence, the correct answer is (d) 9.

Note: We should be aware about the basic expansion because although we can expand algebraic expression raised to the power less than 3 by hand with ease, it becomes much more difficult to expand higher order terms such as in this case for power raised to 7. Thus, in such cases we need to remember the formula for the binomial theorem which can help us to get a particular coefficient value easily.

Complete step-by-step answer:

While solving the problem we should know the basics of binomial theorem and how to expand an algebraic expression raised to the power n. Now, we know that the formula for ${{(a+b)}^{n}}$ is given by $\sum\limits_{r=0}^{n}{\dfrac{n!}{r!\left( n-r \right)!}{{a}^{n-r}}{{b}^{r}}}$. Thus, in general, the ${{r}^{th}}$ from the beginning would be $\dfrac{n!}{(r-1)!\left( n-r+1 \right)!}{{a}^{n-r+1}}{{b}^{r-1}}$. Thus, to solve this problem, we have the ${{7}^{th}}$ term as –

=$\dfrac{n!}{6!\left( n-6 \right)!}{{a}^{n-6}}{{b}^{6}}$

(Now, we know that a =${{2}^{\dfrac{1}{3}}}$ and b = ${{3}^{-\dfrac{1}{3}}}$, since, the expression here is ${{\left( {{2}^{\dfrac{1}{3}}}+{{3}^{-\dfrac{1}{3}}} \right)}^{n}}$)

= \[\dfrac{n!}{6!\left( n-6 \right)!}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{n-6}}{{\left( {{3}^{-\dfrac{1}{3}}} \right)}^{6}}\] -- (1)

Now, the ${{7}^{th}}$ term from the end would be the ${{(n-6)}^{th}}$ term from the beginning. Thus, we have,

= $\dfrac{n!}{\left( n-6 \right)!6!}{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{6}}{{\left( {{3}^{-\dfrac{1}{3}}} \right)}^{n-6}}$ -- (2)

Since, the ratio is 1:6. Thus, we have

$\dfrac{{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{n-6}}{{\left( {{3}^{-\dfrac{1}{3}}} \right)}^{6}}}{{{\left( {{2}^{\dfrac{1}{3}}} \right)}^{6}}{{\left( {{3}^{-\dfrac{1}{3}}} \right)}^{n-6}}}=\dfrac{1}{6}$

Thus,

${{\left( {{2}^{\dfrac{1}{3}}} \right)}^{(n-6)-6}}{{\left( {{3}^{-\dfrac{1}{3}}} \right)}^{6-(n-6)}}=\dfrac{1}{6}$

${{\left( {{2}^{\dfrac{1}{3}}} \right)}^{n-12}}{{\left( {{3}^{-\dfrac{1}{3}}} \right)}^{12-n}}=\dfrac{1}{6}$

${{\left( {{2}^{-\dfrac{1}{3}}} \right)}^{12-n}}{{\left( {{3}^{-\dfrac{1}{3}}} \right)}^{12-n}}=\dfrac{1}{6}$

\[{{\left( {{2}^{-\dfrac{1}{3}}}\times {{3}^{-\dfrac{1}{3}}} \right)}^{12-n}}=\dfrac{1}{6}\]

\[{{\left( {{(2\times 3)}^{-\dfrac{1}{3}}} \right)}^{12-n}}=\dfrac{1}{6}\]

\[{{\left( {{6}^{-\dfrac{1}{3}}} \right)}^{12-n}}=\dfrac{1}{6}\]

\[{{\left( {{6}^{-\dfrac{1}{3}}} \right)}^{12-n}}={{6}^{-1}}\]

Now, we can compare the co-efficient to get the required value of n. We have,

$\dfrac{-(12-n)}{3}=-1$

-(12-n) = -3

12-n = 3

n = 9

Hence, the correct answer is (d) 9.

Note: We should be aware about the basic expansion because although we can expand algebraic expression raised to the power less than 3 by hand with ease, it becomes much more difficult to expand higher order terms such as in this case for power raised to 7. Thus, in such cases we need to remember the formula for the binomial theorem which can help us to get a particular coefficient value easily.

Last updated date: 03rd Jun 2023

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