Question

The sum \( S = \frac{1}{9!} + \frac{1}{3!7!} + \frac{1}{5!5!} + \frac{1}{7!3!} + \frac{1}{9!} \) is equal to:

• A. \( \frac{2^{10}}{8!} \)
• B. \( \frac{2^9}{10!} \)
• C. \( \frac{2^7}{10!} \)
• D. \( \frac{2^6}{10!} \)
• E. \( \frac{2^5}{8!} \)

Hint:

Whenever you see factorials in denominators summing to a constant (here \( 3+7=10, 5+5=10 \)), always try to multiply by that constant's factorial to convert the expression into a binomial sum.

Answer 1

The Correct Option is B

Concept: This series resembles the expansion of \( (1+1)^n \) using binomial coefficients \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \). To simplify the sum, we multiply and divide the entire expression by \( 10! \) to create recognizable binomial terms.

Step 1: Transform the sum into binomial coefficients.
Multiply and divide by \( 10! \): \[ S = \frac{1}{10!} \left[ \frac{10!}{9!1!} + \frac{10!}{3!7!} + \frac{10!}{5!5!} + \frac{10!}{7!3!} + \frac{10!}{9!1!} \right] \] (Note: \( \frac{1}{9!} \) is equivalent to \( \frac{10!}{9!1! \cdot 10} \), but notice the pattern uses only the odd terms \( \binom{10}{1}, \binom{10}{3}, \binom{10}{5}, \dots \)) \[ S = \frac{1}{10!} \left[ \binom{10}{1} + \binom{10}{3} + \binom{10}{5} + \binom{10}{7} + \binom{10}{9} \right] \]

Step 2: Use the property of binomial sums.
The sum of all binomial coefficients is \( \sum_{r=0}^{n} \binom{n}{r} = 2^n \). The sum of only the odd-indexed (or only even-indexed) coefficients is exactly half of the total: \[ \binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \dots = 2^{n-1} \] For \( n = 10 \): \[ \binom{10}{1} + \binom{10}{3} + \binom{10}{5} + \binom{10}{7} + \binom{10}{9} = 2^{10-1} = 2^9 \]

Step 3: Combine the results.
\[ S = \frac{1}{10!} [2^9] = \frac{2^9}{10!} \]