Question

The sum of the rational terms in the binomial expansion of \( \left( \sqrt{2} + 3^{1/5} \right)^{10} \) is:

• A. 41
• B. 39
• C. 32
• D. 30

Hint:

To find rational terms in a binomial expansion, check for integer exponents on the terms. For this problem, \( r \) must be even and divisible by 5 to ensure the exponents are integers.

Answer 1

The Correct Option is A

We are asked to find the sum of the rational terms in the binomial expansion of \( \left( \sqrt{2} + 3^{1/5} \right)^{10} \).
The general term in the binomial expansion of \( (a + b)^n \) is given by: \[ T_{r+1} = \binom{n}{r} a^{n-r} b^r. \] For our expression, \( a = \sqrt{2} \), \( b = 3^{1/5} \), and \( n = 10 \), so the general term is: \[ T_{r+1} = \binom{10}{r} (\sqrt{2})^{10-r} (3^{1/5})^r. \] Simplifying: \[ T_{r+1} = \binom{10}{r} 2^{(10-r)/2} 3^{r/5}. \] 
Step 1: The rational terms in the binomial expansion occur when both exponents of 2 and 3 are integers. Therefore, we need both \( (10 - r)/2 \) and \( r/5 \) to be integers. This implies that:
- \( 10 - r \) must be even, so \( r \) must be even.
- \( r \) must also be divisible by 5.
Step 2: The values of \( r \) that satisfy both conditions are \( r = 0, 10 \). - For \( r = 0 \): \[ T_1 = \binom{10}{0} 2^{10/2} 3^{0} = 1 \times 2^5 = 32. \] - For \( r = 10 \): \[ T_{11} = \binom{10}{10} 2^{0} 3^{2} = 1 \times 1 \times 9 = 9. \] 
Step 3: The sum of the rational terms is the sum of \( T_1 \) and \( T_{11} \): \[ 32 + 9 = 41. \]