Explanation: Solution 1. Identify the Pattern Let the given sum be . Notice that in each term, the sum of the factorials in the denominator is , , etc. To turn these into binomial coefficients, we multiply and divide the entire expression by . 2. Convert to Binomial Coefficients Using the formula for combinations , the expression becomes: 3. Use Binomial Identities We know that for any , the sum of all binomial coefficients is: We also know that the sum of odd indexed binomial coefficients is equal to the sum of even indexed binomial coefficients, which is half of the total sum: 4. Final Calculation In our case, . Therefore, the sum inside the brackets is: Substitute this back into our equation for : Correct Option: (1)

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JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

The value of $\frac{1}{1 ! 50 !} + \frac{1}{3 ! 48 !} + \frac{1}{5 ! 46 !} + \dots + \frac{1}{49 ! 2 !} + \frac{1}{51 ! 0 !}$ is:

Choose the correct answer:

Explanation

Solution

1. Identify the Pattern

Let the given sum be $S$ . Notice that in each term, the sum of the factorials in the denominator is $1+50=51$ , $3+48=51$ , etc. To turn these into binomial coefficients, we multiply and divide the entire expression by $51!$ .

$S = \frac{1}{51!} \left[ \frac{51!}{1!50!} + \frac{51!}{3!48!} + \frac{51!}{5!46!} + \dots + \frac{51!}{51!0!} \right]$

2. Convert to Binomial Coefficients

Using the formula for combinations $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ , the expression becomes:

$S = \frac{1}{51!} \left[ \binom{51}{1} + \binom{51}{3} + \binom{51}{5} + \dots + \binom{51}{51} \right]$

3. Use Binomial Identities

We know that for any $n$ , the sum of all binomial coefficients is:

$\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n} = 2^n$

We also know that the sum of odd indexed binomial coefficients is equal to the sum of even indexed binomial coefficients, which is half of the total sum:

$\binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \dots = 2^{n-1}$

4. Final Calculation

In our case, $n = 51$ . Therefore, the sum inside the brackets is:

$\binom{51}{1} + \binom{51}{3} + \binom{51}{5} + \dots + \binom{51}{51} = 2^{51-1} = 2^{50}$

Substitute this back into our equation for $S$ :

$S = \frac{1}{51!} \times 2^{50}$

$S = \frac{2^{50}}{51!}$

Correct Option: (1)

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