Question

chứng minh (n+1)(n+2)...2n chia hết cho 2ⁿ. tìm thương của phép chia

Answer 1

\(\left(n+1\right)\left(n+2\right)...2n=\frac{\left(2n\right)!}{n!}\)

Ta có: \(\left(2n\right)!=1.2.3.4.....\left(2n-1\right).2n\)\(=\left(2.4.6.8.....2n\right)\left[1.3.5.7....\left(2n-1\right)\right]\)

\(=\left[2.\left(1\right).2.\left(2\right).2.\left(3\right)....2.\left(n\right)\right]\left[1.3.5.7...\left(2n-1\right)\right]\)

\(=2^n.\left(1.2.3.....n\right)\left[1.3.5.7....\left(2n-1\right)\right]\)

\(=2^n.n!.\left[1..3.5...\left(2n-1\right)\right]\)

\(\Rightarrow\frac{\left(2n\right)!}{n!}=2^n.\left[1.3.5.....\left(2n-1\right)\right]\)

Vậy .......

Thương là .......