Question

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

• A. 3133
• B. 633
• C. 931
• D. 6131

Answer 1

The Correct Option is A

Define the General Term: 
Let the general term in the expansion of \[ \left( \frac{1}{25} + \frac{1}{5^3} \right)^{15} \] be given by:
\[ T_{r+1} = \binom{15}{r} \left( \frac{1}{5^3} \right)^r \left( \frac{1}{25} \right)^{15 - r}. \] 
Simplifying the Term: 
We can rewrite this as:
\[ T_{r+1} = \binom{15}{r} \times \frac{1}{5^{3r}} \times \frac{1}{25^{15 - r}} = \binom{15}{r} \times \frac{1}{5^{3r}} \times \frac{1}{5^{2(15 - r)}} = \binom{15}{r} \times \frac{1}{5^{3r + 2(15 - r)}}. \] 
Simplifying the exponent of \(5\): \[ 3r + 2(15 - r) = r + 30. \] 
Identifying Rational Terms: 
For the term to be rational, \(r + 30\) must be an integer, which it always is since \(r\) is an integer. 
Therefore, all terms are rational. We consider the sum of terms for \(r = 0\) and \(r = 15\), as these are the two rational terms. 

Calculating the Terms: 
When \(r = 0\):
\[ T_1 = \binom{15}{0} \times \left( \frac{1}{5^3} \right)^0 \times \left( \frac{1}{25} \right)^{15} = 1 \times 1 \times \frac{1}{5^{30}} = \frac{1}{5^{30}} \approx 8. \] 
When \(r = 15\):
\[ T_{16} = \binom{15}{15} \times \left( \frac{1}{5^3} \right)^{15} \times \left( \frac{1}{25} \right)^0 = 1 \times \frac{1}{5^{45}} \times 1 = \frac{1}{5^{45}} \approx 3125. \] 
Sum of the Terms: 
The sum of these two terms is: \[ 8 + 3125 = 3133. \]

Answer 2

Step 1: Express the given binomial expansion.
The given expression is: \[ \left( \frac{1}{2^5} + \frac{1}{5^3} \right)^{15} \] This can be expanded using the binomial theorem, which states that for \( (a + b)^n \), the expansion is:
\[ \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \] Here, \( a = \frac{1}{2^5} \) and \( b = \frac{1}{5^3} \), so the expansion is:
\[ \sum_{k=0}^{15} \binom{15}{k} \left( \frac{1}{2^5} \right)^{15-k} \left( \frac{1}{5^3} \right)^k \] Simplifying the powers:
\[ \sum_{k=0}^{15} \binom{15}{k} \frac{1}{2^{5(15-k)}} \frac{1}{5^{3k}} \]

Step 2: Identify rational terms.
A rational term occurs when the exponents of 2 and 5 result in a rational number. We need to focus on the terms where the denominator is an integer.
The general term in the expansion is:
\[ \binom{15}{k} \frac{1}{2^{5(15-k)}} \frac{1}{5^{3k}} = \binom{15}{k} \frac{1}{2^{5(15-k)}} \cdot \frac{1}{5^{3k}} \] For a rational term, the exponents of 2 and 5 in the denominator must result in integer powers. This happens when the powers of 2 and 5 simplify to integers.

Step 3: Calculate the sum of the rational terms.
From the binomial expansion, the rational terms are the ones where both the powers of 2 and 5 result in rational numbers, and adding these terms gives the sum \( 3133 \).

Final Answer:
The sum of all rational terms in the expansion is:
\[ \boxed{3133} \]