Math - product of AP

咁簡單嘅數都用MI好似唔係咁好喎！


Prove (n + 1)(n + 2) ...… (2n) = 2ⁿ × 1 × 3 × 5 × …… × (2n – 1).

　L.H.S.
= (n + 1)(n + 2) ...… (2n)
= (1 × 2 × 3 × …… n(n + 1)(n + 2) ...… (2n))/(1 × 2 × 3 × …… n)
= [(2 × 4 × 6 × …… (2n))(1 × 3 × 5 × …… (2n – 1))]/(1 × 2 × 3 × …… n)
= [2ⁿ (1 × 2 × 3 × …… n)(1 × 3 × 5 × …… (2n – 1))]/(1 × 2 × 3 × …… n)
= 2ⁿ × 1 × 3 × 5 × …… × (2n – 1)
= R.H.S.

∴(n + 1)(n + 2) ...… (2n) = 2ⁿ × 1 × 3 × 5 × …… × (2n – 1)

Using MI, we can prove the result
let S(n) be the statement " (n+1)(n+2).....(2n)= 2^n x 1 x 3 x ......x (2n-1) for all positive integers n."

when n=1
LHS=2
RHS=2^1*1=2

so, when n=1 the statement is true

Assume when n=k, the statement is true that is
(k+1)(k+2).....(2k)= 2^k x 1 x 3 x ......x (2k-1)
when n=k+1
LHS
=(k+2).....(2k)(2k+1)(2k+2)
=2^k x 1 x 3 x ......x (2k-1)x (2k+1) x (2k+2)/(k+1) [using assumption]
=2^(k+1) x 1 x 3 x ......x (2k-1)x (2k+1) x (k+1)/(k+1)
=2^(k+1) x 1 x 3 x ......x (2k-1)x (2k+1)
=RHS
So,when n=k+1, the statement is true
By MI for all positive values of n, S(n) is true