Question

Let E = (2n + 1) (2n + 3) (2n + 5)…(4n – 3) (4n – 1) where n > 1. then 2ⁿ Eis divisible by

• A. ²ⁿC_n
• B. ⁴ⁿC_2n
• C. ²ⁿC_n/2
• D. ⁴ⁿC_n/2

Answer 1

Answer: (B)

⁴ⁿC_2n

Answer 2

E = (2n +1) (2n + 3) (2n + 5)……(4n – 3) (4n – 1)

E =$\frac{(2n)!(2n + 1)(2n + 2)(2n + 3)(2n + 4)(2n + 5)....(4n - 1)4n}{(2n)!(2n + 2)(2n + 4)....(4n)}$

E = $\frac{(4n)!n!}{(2n)!n!2^{n}(n + 1)(n + 2).....(2n)}$

E = $\frac{(4n)!n!}{(2n)!(2n)!2^{n}}$ ̃ 2ⁿ E = n! ⁴ⁿC_2n

Hence 2ⁿE is divisible by ⁴ⁿC_2n