Number of integral terms in the expansion of {7^((1/2))+11^((1/6))}⁸²⁴ is equal to _________.

Solution

Maths sir · 4 steps

Write the general term, then make both powers whole numbers; that is a condition on r alone.

1. General term. The (r+1)th term of the expansion is written below, for r = 0, 1, 2, …, 824. The binomial coefficient is always an integer, so only the two powers matter.

   Tᵣ₊₁ = C(824, r) 7^((824-r)/2) 11^(r/6)

2. Conditions on r. r/6 must be a whole number, so r is a multiple of 6. Also (824-r)/2 must be a whole number. Since 824 is even, this needs r even, and every multiple of 6 is already even.
3. Count the values. So r = 0, 6, 12, …, 822, because 822 is the largest multiple of 6 not above 824. That is 822/6 + 1 = 137 + 1 values, where the extra 1 counts r = 0.
4. So. The expansion has 138 integral terms.

The idea to keep. For integral terms, make every exponent in Tᵣ₊₁ a whole number, combine the conditions on r (usually through an LCM), then count the values of r from 0 to n, including r = 0.

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