`#Rùa`
Đáp án+Giải thích các bước giải:
Ta có : `1*3*5*...*(2n-1)`
`=[[1*3*5*...*(2n-1)](2*4*6*....*2n)]/(2*4*6*...*2n)`
`=[(1*2*3*4*5*6*....*n)[(n+1)(n+2)....(2n-1)2n]]/[(1*2)(2*2)(2*3)...(2*n)]`
`=[(1*2*3*4*5*6*....*n)[(n+1)(n+2)....(2n-1)2n]]/[(1*2*3*....*n)\underbrace{(2*2*2*....*2)}_{n số 2}]`
`=[(n+1)(n+2)....(2n-1)2n]/(2^n)`
`-> (1*3*5*...(2n-1))/[(n+1)(n+2)....(2n-1)2n]`
`={[(n+1)(n+2)....(2n-1)2n]/(2^n)}/[(n+1)(n+2)....(2n-1)2n]`
`=[(n+1)(n+2)....(2n-1)2n]/(2^n) . 1/[(n+1)(n+2)....(2n-1)2n]`
`=1/(2^n)`
Vậy `(1*3*5*...(2n-1))/[(n+1)(n+2)....(2n-1)2n]=1/(2^n)`
