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 :
Solution
$$
\begin{aligned}
& \mathrm{S}=\frac{1}{1 ! 50 !}+\frac{1}{3 ! 48 !}+\frac{1}{5 ! 46 !}+\ldots \ldots+\frac{1}{49 ! 2 !}+\frac{1}{51 ! 1 !} \\\\
& =\frac{1}{51 !}\left(\frac{51 !}{1 ! 50 !}+\frac{51 !}{3 ! 48 !}+\frac{51 !}{5 ! 46 !}+\ldots . .+\frac{51 !}{49 ! 2 !}+\frac{51 !}{51 ! 0 !}\right) \\\\
& =\frac{1}{51 !}\left({ }^{51} C_{50}+{ }^{51} C_{48}+{ }^{51} C_{46}+\ldots \ldots .+{ }^{51} C_2+{ }^{51} C_0\right) \\\\
& \because{ }^n C_0+{ }^n C_2+{ }^n C_4+\ldots \ldots=2^{n-1} \\\\
& \therefore S=\frac{2^{50}}{51 !}
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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