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Question: Find the relation between r and n in order that the coefficients of the \({\left( {3r} \right)^{th}}...

Find the relation between r and n in order that the coefficients of the (3r)th{\left( {3r} \right)^{th}}and (r+2)th{\left( {r + 2} \right)^{th}}term of (1+x)2n{\left( {1 + x} \right)^{2n}}may be equal.
(a) r = n (b) 2r = n (c) r = - n (d) r = - 4n  (a){\text{ r = n}} \\\ (b){\text{ 2r = n}} \\\ (c){\text{ r = - n}} \\\ (d){\text{ r = - 4n}} \\\

Explanation

Solution

Hint: In this question it is given that the (3r)th{\left( {3r} \right)^{th}} and the (r+2)th{\left( {r + 2} \right)^{th}} term in expansion of (1+x)2n{\left( {1 + x} \right)^{2n}}is equal. Thus use the formula for any general term which is Tr+1=2nCr(xr){{\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
(1+x)2n{\left( {1 + x} \right)^{2n}}
Now it is given that the (3r)th{\left( {3r} \right)^{th}}and (r+2)th{\left( {r + 2} \right)^{th}} term in the expansion of (1+x)2n{\left( {1 + x} \right)^{2n}} is equal.
Now as we know the general term(i.e.(r+1)th term)\left( {{\text{i}}{\text{.e}}{\text{.}}{{\left( {r + 1} \right)}^{th}}{\text{ term}}} \right)in the expansion of (1+x)2n{\left( {1 + x} \right)^{2n}} is given as
Tr+1=2nCr(xr){{\text{T}}_{r + 1}} = {}^{2n}{C_r}\left( {{x^r}} \right)
So, the coefficient of (3r)th{\left( {3r} \right)^{th}} in the expansion of (1+x)2n{\left( {1 + x} \right)^{2n}} is equal to
⇒2nC3r−1\Rightarrow {}^{2n}{C_{3r - 1}}
And the coefficient of (r+2)th{\left( {r + 2} \right)^{th}} in the expansion of (1+x)2n{\left( {1 + x} \right)^{2n}} is equal to
⇒2nCr+1\Rightarrow {}^{2n}{C_{r + 1}}
Now both the coefficients are equal according to given condition
⇒2nC3r−1=2nCr+1\Rightarrow {}^{2n}{C_{3r - 1}} = {}^{2n}{C_{r + 1}}
Now as we know ⇒nCr=n!r!(n−r)! \Rightarrow {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} so, use this property in above equation we have,
⇒2n!(3r−1)!(2n−3r+1)!=2n!(r+1)!(2n−r−1)!\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
(i)3r−1=r+1, 2n−3r+1=2n−r−1\left( i \right)3r - 1 = r + 1,{\text{ }}2n - 3r + 1 = 2n - r - 1
⇒2r=2, 2r=2\Rightarrow 2r = 2,{\text{ }}2r = 2
⇒r=1\Rightarrow r = 1
(ii)3r−1=2n−r−1, 2n−3r+1=r+1\left( {ii} \right)3r - 1 = 2n - r - 1,{\text{ }}2n - 3r + 1 = r + 1
⇒4r=2n\Rightarrow 4r = 2n
⇒2r=n\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.