Let S = 1 25 + 1 3 ? 23 + 1 5 ? 21 + ? up to 13 terms. If 13 S = 2 k n , k , then n + k is equal to [2026]

Question

Let $S = \frac{1}{25!} + \frac{1}{3! 23!} + \frac{1}{5! 21!} + \dots$ up to 13 terms. If $13 S = \frac{2^{k}}{n!}, k \in ℕ,$ then $n + k$ is equal to [2026]

Smart Achievers · Accepted Answer

(2)

$\frac{1}{26!} (\frac{26!}{25! 1!} + \frac{26!}{3! 23!} + \frac{26!}{5! 21!} + \dots + 13 terms)$

$= \frac{1}{26!} ({}^{26}C_{1} + {}^{26}C_{3} + {}^{26}C_{5} + \dots + 13 terms)$

$= \frac{1}{26!} ({}^{26}C_{1} + {}^{26}C_{3} + {}^{26}C_{5} + \dots + {}^{26}C_{25})$

$\Rightarrow S = \frac{1}{26!} \times 2^{25}$

$\Rightarrow 13 S = \frac{2^{24}}{25!}$

$so n + k = 25 + 24 = 49$

Solution

Q.
Let $S = \frac{1}{25!} + \frac{1}{3! 23!} + \frac{1}{5! 21!} + \dots$ up to 13 terms. If $13 S = \frac{2^{k}}{n!}, k \in ℕ,$ then $n + k$ is equal to [2026]

1 50

2 49

3 52

4 51

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