Question

The number of integral terms in the expansion of $ \left( 5^{\frac{1}{2}} + 7^{\frac{1}{8}} \right)^{1016} $ is:

• A. 130
• B. 128
• C. 127
• D. 129

Hint:

For binomial expansions involving fractional exponents, ensure that the exponents of the terms are integers for the terms to be integral. This can be done by ensuring that the powers of the terms satisfy the divisibility conditions.

Answer 1

The Correct Option is B

To determine the count of integral terms in the expansion of \( \left( 5^{\frac{1}{2}} + 7^{\frac{1}{8}} \right)^{1016} \), the binomial theorem is applied. The general term of \( (a + b)^n \) is \(T_{k+1} = \binom{n}{k} a^{n-k} b^k\). For this expression, \( a = 5^{\frac{1}{2}} \) and \( b = 7^{\frac{1}{8}} \). The general term is therefore:

\(T_{k+1} = \binom{1016}{k} (5^{\frac{1}{2}})^{1016-k} (7^{\frac{1}{8}})^k\)

This simplifies to:

\(T_{k+1} = \binom{1016}{k} 5^{\frac{1016-k}{2}} 7^{\frac{k}{8}}\)

For the term to be integral, the exponents of both 5 and 7 must be integers. This requires:

• \(\frac{1016-k}{2}\) to be an integer, meaning \( 1016-k \) must be even.
• \(\frac{k}{8}\) to be an integer, meaning \( k \) must be a multiple of 8.

Since \( 1016 \) is even, any \( k \) that is a multiple of 8 will also make \( 1016-k \) even, satisfying both conditions. The possible values for \( k \) are multiples of 8 within the range of the binomial expansion, from 0 to 1016.

The sequence of valid \( k \) values is:

\( k = 0, 8, 16, \dots, 1016 \)

This arithmetic sequence has a first term of 0, a common difference of 8, and a last term of 1016. The number of terms (\( n \)) is found using the formula \( k = a + (n-1)d \):

\(1016 = 0 + (n-1) \cdot 8\)

Solving for \( n \):

\(n-1 = \frac{1016}{8} = 127 \Rightarrow n = 128\)

There are 128 integral terms.