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

Answer: (A)

3133

Answer 2

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: $8 + 3125 = 3133.$