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In the binomial expansion of (7^(1/3) + 11^(1/9))⁶⁵⁶¹, find the number of terms that are free from radicals (i.e., terms in which both 7 and 11 appear with integer exponents).
- 730
- 729
- 725
- 750
Correct answer: 730
Solution
r must be a multiple of 9 (giving 730 values: r = 0, 9, 18,..., 6561). Since 6561 = 3⁸ is divisible by 3, and any multiple of 9 is also a multiple of 3, the condition (6561 - r) divisible by 3 is automatically satisfied for all such r. Hence all 730 terms qualify.
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