# Find the relation between r and n in order that the coefficients of the ${\left( {3r} \right)^{th}}$and ${\left( {r + 2} \right)^{th}}$term of ${\left( {1 + x} \right)^{2n}}$may be equal.

$

(a){\text{ r = n}} \\

(b){\text{ 2r = n}} \\

(c){\text{ r = - n}} \\

(d){\text{ r = - 4n}} \\

$

Last updated date: 27th Mar 2023

•

Total views: 307.2k

•

Views today: 6.84k

Answer

Verified

307.2k+ views

Hint: In this question it is given that the ${\left( {3r} \right)^{th}}$ and the ${\left( {r + 2} \right)^{th}}$ term in expansion of ${\left( {1 + x} \right)^{2n}}$is equal. Thus use the formula for any general term which is \[{{\text{T}}_{r + 1}} = {}^{2n}{C_r}\left( {{x^r}} \right)\] for the expansion of given entity to obtain the relation between r and n.

Complete step-by-step answer:

Given equation is

${\left( {1 + x} \right)^{2n}}$

Now it is given that the ${\left( {3r} \right)^{th}}$and ${\left( {r + 2} \right)^{th}}$ term in the expansion of ${\left( {1 + x} \right)^{2n}}$ is equal.

Now as we know the general term\[\left( {{\text{i}}{\text{.e}}{\text{.}}{{\left( {r + 1} \right)}^{th}}{\text{ term}}} \right)\]in the expansion of \[{\left( {1 + x} \right)^{2n}}\] is given as

\[{{\text{T}}_{r + 1}} = {}^{2n}{C_r}\left( {{x^r}} \right)\]

So, the coefficient of ${\left( {3r} \right)^{th}}$ in the expansion of ${\left( {1 + x} \right)^{2n}}$ is equal to

$ \Rightarrow {}^{2n}{C_{3r - 1}}$

And the coefficient of ${\left( {r + 2} \right)^{th}}$ in the expansion of ${\left( {1 + x} \right)^{2n}}$ is equal to

$ \Rightarrow {}^{2n}{C_{r + 1}}$

Now both the coefficients are equal according to given condition

$ \Rightarrow {}^{2n}{C_{3r - 1}} = {}^{2n}{C_{r + 1}}$

Now as we know $ \Rightarrow {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so, use this property in above equation we have,

$ \Rightarrow \dfrac{{2n!}}{{\left( {3r - 1} \right)!\left( {2n - 3r + 1} \right)!}} = \dfrac{{2n!}}{{\left( {r + 1} \right)!\left( {2n - r - 1} \right)!}}$

Now comparing the denominator terms as numerator is equal we have

$\left( i \right)3r - 1 = r + 1,{\text{ }}2n - 3r + 1 = 2n - r - 1$

$ \Rightarrow 2r = 2,{\text{ }}2r = 2$

$ \Rightarrow r = 1$

$\left( {ii} \right)3r - 1 = 2n - r - 1,{\text{ }}2n - 3r + 1 = r + 1$

$ \Rightarrow 4r = 2n$

$ \Rightarrow 2r = n$

Hence option (b) is correct.

Note: Whenever we face such a type of problem the key concept is simply to have the gist of any general term in the series expansion. This formula based approach will eventually help you get on the right track to reach the desired relation between the variables involved in this expansion.

Complete step-by-step answer:

Given equation is

${\left( {1 + x} \right)^{2n}}$

Now it is given that the ${\left( {3r} \right)^{th}}$and ${\left( {r + 2} \right)^{th}}$ term in the expansion of ${\left( {1 + x} \right)^{2n}}$ is equal.

Now as we know the general term\[\left( {{\text{i}}{\text{.e}}{\text{.}}{{\left( {r + 1} \right)}^{th}}{\text{ term}}} \right)\]in the expansion of \[{\left( {1 + x} \right)^{2n}}\] is given as

\[{{\text{T}}_{r + 1}} = {}^{2n}{C_r}\left( {{x^r}} \right)\]

So, the coefficient of ${\left( {3r} \right)^{th}}$ in the expansion of ${\left( {1 + x} \right)^{2n}}$ is equal to

$ \Rightarrow {}^{2n}{C_{3r - 1}}$

And the coefficient of ${\left( {r + 2} \right)^{th}}$ in the expansion of ${\left( {1 + x} \right)^{2n}}$ is equal to

$ \Rightarrow {}^{2n}{C_{r + 1}}$

Now both the coefficients are equal according to given condition

$ \Rightarrow {}^{2n}{C_{3r - 1}} = {}^{2n}{C_{r + 1}}$

Now as we know $ \Rightarrow {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so, use this property in above equation we have,

$ \Rightarrow \dfrac{{2n!}}{{\left( {3r - 1} \right)!\left( {2n - 3r + 1} \right)!}} = \dfrac{{2n!}}{{\left( {r + 1} \right)!\left( {2n - r - 1} \right)!}}$

Now comparing the denominator terms as numerator is equal we have

$\left( i \right)3r - 1 = r + 1,{\text{ }}2n - 3r + 1 = 2n - r - 1$

$ \Rightarrow 2r = 2,{\text{ }}2r = 2$

$ \Rightarrow r = 1$

$\left( {ii} \right)3r - 1 = 2n - r - 1,{\text{ }}2n - 3r + 1 = r + 1$

$ \Rightarrow 4r = 2n$

$ \Rightarrow 2r = n$

Hence option (b) is correct.

Note: Whenever we face such a type of problem the key concept is simply to have the gist of any general term in the series expansion. This formula based approach will eventually help you get on the right track to reach the desired relation between the variables involved in this expansion.

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