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Question
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
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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
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