The sum of all rational terms in the expansion of (1+2^(1 / 3)+3^(1 / 2))⁶ is equal to _________.

Solution

Maths sir · 3 steps

Write the general multinomial term, then keep only powers that make both surds whole numbers.

1. General term. Take 1 a total of a times, 2^(1/3) a total of b times and 3^(1/2) a total of c times, with a + b + c = 6. Each such choice gives the term below.

   T = 6!/(a! b! c!) 2^(b/3) 3^(c/2)

2. Rational condition. 2^(b/3) is rational only when b is a multiple of 3, so b ∈ {0, 3, 6}. 3^(c/2) is rational only when c is even, so c ∈ {0, 2, 4, 6}. We also need b + c ≤ 6.

   (a, b, c)	6!/(a! b! c!)	2^(b/3) 3^(c/2)	Term
   (6, 0, 0)	1	1	1
   (4, 0, 2)	15	3	45
   (2, 0, 4)	15	9	135
   (0, 0, 6)	1	27	27
   (3, 3, 0)	20	2	40
   (1, 3, 2)	60	2 × 3 = 6	360
   (0, 6, 0)	1	4	4

3. So. The sum is 1 + 45 + 135 + 27 + 40 + 360 + 4 = 612.

The common slip: 252. Leaving out the mixed term (a, b, c) = (1, 3, 2), which uses both surds at once, drops 360 and gives 252.

The idea to keep. For rational terms, make each surd's exponent a whole number: the power of p^(1/m) must be a multiple of m. Do not forget the mixed terms where both surds appear together.

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