Question

The number of integral terms in the binomial expansion of \[ \left(11^{\frac{1}{2}} + 17^{\frac{1}{8}}\right)^{1024} \] is:

• A. \(129\)
• B. \(131\)
• C. \(133\)
• D. \(137\)

Hint:

In binomial expansions with radicals, convert all exponents to a common denominator and use modular arithmetic to count integral terms efficiently.

Answer 1

The Correct Option is A

Concept:
The general term of \( (a+b)^n \) is \[ T_{k+1} = \binom{n}{k} a^{n-k} b^k \]
A term is integral if the combined exponent of all radicals is an integer.
Congruence conditions help count valid values of \(k\).
Step 1: Write the general term \[ T_{k+1} = \binom{1024}{k} \left(11^{\frac{1}{2}}\right)^{1024-k} \left(17^{\frac{1}{8}}\right)^k \]
Step 2: Simplify the powers \[ = \binom{1024}{k} \, 11^{\frac{1024-k}{2}} \, 17^{\frac{k}{8}} \] Write everything with denominator \(8\): \[ 11^{\frac{4(1024-k)}{8}} \cdot 17^{\frac{k}{8}} \] The total power of radicals is: \[ \frac{4096 - 3k}{8} \]
Step 3: Condition for integrality For the term to be integral: \[ 4096 - 3k \equiv 0 \pmod{8} \] \[ 3k \equiv 0 \pmod{8} \] Since \(3\) is coprime with \(8\), \[ k \equiv 0 \pmod{8} \]
Step 4: Count valid values of \(k\) \[ k = 0, 8, 16, \ldots, 1024 \] Number of such values: \[ \frac{1024}{8} + 1 = 128 + 1 = 129 \]