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
327k+ views
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
•
Total views: 327k
•
Views today: 5.87k
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE
