$\frac{1}{1 !(n-1) !}+\frac{1}{3 !(n-3) !}+\frac{1}{5 !(n-5) !}+\ldots$ is equal to

(A) $\frac{2^n}{n !}$

(B) $\frac{2^{n-1}}{n !}$

(C) 0

(D) None of these

0 Votes

Best answer

**SOLUTION —**

$\begin{array}{l}\text { }\frac{1}{1 !(n-1) !}+\frac{1}{3 !(n-3) !}+\frac{1}{5 !(n-5) !}+\ldots \\=\frac{1}{n !}\left[\frac{1}{1 !} \cdot \frac{n !}{(n-1) !}+\frac{1}{3 !} \cdot \frac{n !}{(n-3) !}+\frac{1}{5 !} \cdot \frac{n !}{(n-5) !}+\ldots\right] \\=\frac{1}{n !}\left[{ }^n C_1+{ }^n C_3+{ }^n C_5+\ldots\right] \\=\frac{2^{n-1}}{n !} \\\end{array}$

So, The correct option will be **(B).**

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