Question

CMR:

51/2 . 52/2 . 53/2 . ... . 100/2 = 1 . 3 . 5 . ... . 99

Answer 1

Đặt C=1.3.5.7...99
Đặt D=51/2.52/2.53/2 ....100/2
Ta có:C=1.3.5.7...99
=>2.4.6...100.C=1.2.3...100
=>C = (1.2.3....100) / (2.4.6...100)= (1.2.3...50).(51.52...100) / [(2.1)(2.2).(2.3)...(2.50)]
C=(1.2.3...50).(51.52...100) /[2^50.(1.2.3...50)] =(51.52...100)/2^50 =51/2.52/2.53/2...100/2 =D
Vậy C=D

Answer 2

Ta có :

\(1.3.5.....99=\frac{\left(1.3.5.....99\right)\left(2.4.6.....98\right)}{2.4.6.....98}=\frac{1.2.3.....99.100}{2^{50}\left(1.2.3.....50\right)}=\frac{51.52.53.....100}{2.2.2.....2}\)

                       \(=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}.....\frac{100}{2}\)

Vậy......................

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