# Given the integers r > 1, n > 2, and coefficients of (3r)th and (r + 2)nd terms in the binomial expansion of (1 + x)2n are equal, then ______. - Mathematics

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Given the integers r > 1, n > 2, and coefficients of (3r)th and (r + 2)nd terms in the binomial expansion of (1 + x)2n are equal, then ______.

#### Options

• n = 2r

• n = 3r

• n = 2r + 1

• None of these

#### Solution

Given the integers r > 1, n > 2, and coefficients of (3r)th and (r + 2)nd terms in the binomial expansion of (1 + x)2n are equal, then n = 2r.

Explanation:

Given that r > 1 and n > 2

Then "T"_(3r) = "T"_(3r - 1 + 1)

= ""^(2n)"C"_(3r - 1) * x^(3r - 1)

And "T_(r + 2) = "T"_(r + 1 + 1)

= ""^(2n)"C"_(r + 1) x^(r + 1)

We have ""^(2n)"C"_(3r - 1) = ""^(2n)"C"_(r + 1)

⇒ 3r – 1 + r + 1 = 2n   ....[because ""^n"C"_p = ""^n"C"_q ⇒ n = p + q]

⇒ 4r = 2n

n = 2r

Concept: Binomial Theorem for Positive Integral Indices
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#### APPEARS IN

NCERT Mathematics Exemplar Class 11
Chapter 8 Binomial Theorem
Exercise | Q 19 | Page 144
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