Question

The sum of the series $S = \frac{1}{19! } + \frac{1}{3!7!} + \frac{1}{5! 5!} + ... $ to 10 terms is equal to :

• A. $\frac{2^{20}}{20!}$
• B. $\frac{2^{10}}{20!}$
• C. $\frac{2^{19}}{19!}$
• D. $\frac{2^{19}}{20!}$

Answer 1

The Correct Option is B

To solve the problem of finding the sum of the given series, we need to first understand the structure of the series:

The series given is:

S = \frac{1}{19! } + \frac{1}{3!7!} + \frac{1}{5! 5!} + \ldots

Upon observation, we notice that each term of the series can be represented in the form of \frac{1}{(2n-1)!(20-2n)!} for n = 1, 2, ..., 10.

This corresponds to the expansion of 1!\cdot19! + 3!\cdot17! + 5!\cdot15! + \ldots in the binomial theorem when finding the coefficient of x^{10} in the expansion of (1 + x)^{20}.

The general term in the binomial expansion of (1 + x)^{20} is given by:

\binom{20}{k} \cdot x^k

Here, this translates to the coefficients of terms involving factorials that are present in our series.

Let's calculate the sum of these coefficients using the known property of binomial coefficients:

\binom{20}{0} + \binom{20}{1} + \binom{20}{2} + \ldots + \binom{20}{10} = 2^{20-1} = 2^{19}

However, the derived series has alternating factorials in terms of even numbers, consuming half of the series:

Thus, the sum of the relevant series portion:

\sum \text{of terms involving odd factorials from 0 to 10} = \frac{2^{10}}{20!}

Hence, the sum of the series to 10 terms is:

\frac{2^{10}}{20!}

Thus, the correct option is:

\frac{2^{10}}{20!}