JEE Mains · Maths · STD 11 - 7. binomial theoram
The value of \(\frac{1}{1 ! 50 !}+\frac{1}{3 ! 48 !}+\frac{1}{5 ! 46 !}+\ldots .+\frac{1}{49 ! 2 !}+\frac{1}{51 ! 1 !}\) is \(.............\).
- A \(\frac{2^{50}}{50 !}\)
- B \(\frac{2^{50}}{51 !}\)
- C \(\frac{2^{51}}{51 !}\)
- D \(\frac{2^{51}}{50 \text { ! }}\)
Answer & Solution
Correct Answer
(B) \(\frac{2^{50}}{51 !}\)
Step-by-step Solution
Detailed explanation
\(\sum \limits_{ r =1}^{26} \frac{1}{(2 r -1) !(51-(2 r -1)) !}=\sum \limits_{ r =1}^{26}{ }^{51} C _{(2 r -1)} \frac{1}{51 !}\) \(=\frac{1}{51 !}\left\{{ }^{51} C _1+{ }^{51} C _3+\ldots .+{ }^{51} C _{51}\right\}\) \(=\frac{1}{51 !}\left(2^{50}\right)\)
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