The least value of n for which the number of integral terms in the Binomial expansion of (∛7+¹²√11)ⁿ is 183, is :

Solution

Maths sir · 4 steps

Both powers must be integers, so r must be a multiple of 12 and n must be a multiple of 3.

1. General term. Tᵣ₊₁ = C(n, r) 7^((n-r)/3) 11^(r/12). This term is an integer only when both exponents are whole numbers.
2. Conditions on r. r/12 must be an integer, so r = 0, 12, 24, …. Then r is also a multiple of 3, so (n-r)/3 is an integer only if n is a multiple of 3.
3. Count the terms. The allowed values r = 0, 12, 24, … up to n number ⌊ n/12 ⌋ + 1. Do not forget r = 0.

   ⌊ n/12 ⌋ + 1 = 183 ⇒ ⌊ n/12 ⌋ = 182 ⇒ 2184 ≤ n ≤ 2195

   n	⌊ n/12 ⌋	Integral terms
   2148	179	180
   2172	181	182
   2184	182	183
   2196	183	184

4. So. The smallest n in 2184 ≤ n ≤ 2195 is 2184 = 3 × 728, which is also a multiple of 3. The least value is n = 2184, option (a).

The idea to keep. For integral terms, make every exponent in Tᵣ₊₁ a whole number. Then count the allowed values of r from 0 to n, and remember that r = 0 counts too.

Answer from the NTA final answer key. Solution by PrepWiser.