1.If n∈Nn \in \mathbb{N}n∈N and nnn is even, then 11!(n−1)!+13!(n−3)!+15!(n−5)!+⋯+1(n−1)!1!\frac{1}{1!(n-1)!} + \frac{1}{3!(n-3)!} + \frac{1}{5!(n-5)!} + \cdots + \frac{1}{(n-1)!1!}1!(n−1)!1+3!(n−3)!1+5!(n−5)!1+⋯+(n−1)!1!1 equals:a.2nn!\frac{2^n}{n!}n!2nb.2n−1n!\frac{2^{n-1}}{n!}n!2n−1c.2nn!2^n n!2nn!d.None of theseLogin to continueOnly logged in users canattempt or see the solution.
