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JEE Mathematics 2024 Question with Solution

The sum of all rational terms in the expansion of (125+153)15\left(\frac{1}{2^5} + \frac{1}{5^3}\right)^{15} is equal to:

  • A

    31333133

  • B

    633633

  • C

    931931

  • D

    61316131

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Find the sum of all rational terms in the expansion of (125+153)15\left(\frac{1}{2^5} + \frac{1}{5^3}\right)^{15}.

Find: The numerical sum of all rational terms.

The solution is unrelated to the given question and discusses coefficients of powers of xx in another expansion. Hence, it cannot be used to derive the answer for this question.

Using the source answer key, the correct option is A, which corresponds to 31333133.

Common mistakes

  • Treating every term in the binomial expansion as irrational. Here, one must check which powers produce rational values before summing them.

  • Confusing the exponents on 22 and 55 inside the terms. The expressions are 125\frac{1}{2^5} and 153\frac{1}{5^3}, so careless rewriting changes the expansion.

  • Adding coefficients instead of adding the actual rational terms. The question asks for the sum of rational terms, not the sum of binomial coefficients.

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