Given the integers r > 1, n > 2, and coefficient of (3r)^th and (r + 2)^nd terms in the binomial expansion of (1 + x)²ⁿ are equal, then
(a) n = 2r
(b) n = 3r
(c) n = 2r + 1
(d) none of these

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Solution

Given r > 1 and n > 2
and coefficient of T_3r = coefficient of T_r+2 is expansion of (1 + x)²ⁿ
$i . e .^{2 n} C_{3 r - 1} =^{2 n} C_{r + 1} i . e . 3 r - 1 = r + 1 (using property of combination) and 2 n - 3 r + 1 = r + 1 i . e . 2 r = 2 and 2 n = 4 r i . e . r = 1 and n = 2 r$

Hence, the correct answer is option A.

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Given the integers r > 1, n > 2, and coefficient of (3r)th and (r + 2)nd terms in the binomial expansion of (1 + x)2n are equal, then (a) n = 2r (b) n = 3r (c) n = 2r + 1 (d) none of these

Given r > 1 and n > 2 and coefficient of T3r = coefficient of Tr+2 is expansion of (1 + x)2n i.e. 2nC3r-1=2nCr+1i.e. 3r-1=r+1 (using property of combination)and 2n-3r+1=r+1i.e. 2r=2 and 2n=4ri.e. r=1 and n=2r Hence, the correct answer is option A.

Given the integers r > 1, n > 2, and coefficient of (3r)^th and (r + 2)^nd terms in the binomial expansion of (1 + x)²ⁿ are equal, then
(a) n = 2r
(b) n = 3r
(c) n = 2r + 1
(d) none of these