Question

tìm n để \(\frac{1.3.5.....\left\{2n-1\right\}}{\left\{n+1\right\}.\left\{n+2\right\}.....2n}\)= \(\frac{1}{2^n}\). với n \(\varepsilonℕ^∗\)

Answer 1

CÓ THỂ LÀ RẤT KHÓ

Answer 2

ko phải khó mà rất khó

Answer 3

\(\frac{1\cdot3\cdot5\cdot\cdot\cdot\cdot\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\cdot\cdot\cdot2n}=\frac{\left[\left(1\cdot3\cdot5\cdot\cdot\cdot\cdot\left(2n-1\right)\right)\right]\left(2\cdot4\cdot6\cdot\cdot\cdot\cdot2n\right)}{\left(n+1\right)\left(n+2\right)\cdot\cdot\cdot2n\left(2\cdot4\cdot6\cdot\cdot\cdot2n\right)}\)

\(=\frac{1\cdot2\cdot3\cdot4\cdot5\cdot\cdot\cdot\cdot\left(2n-1\right)\cdot2n}{2^n\left(1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot n\right)\left(n+1\right)\left(n+2\right)\cdot\cdot\cdot2n}\)

\(=\frac{1}{2^n}\)