[1+1/2][1+1/3][1+1/4][1+1/5]...[1+1/n]=11 berapakah ?

Perhatikan!
U1 → [1+½] = 3/2 = 1,5
U2 → [1+½][1+⅓] = (3/2)(4/3) = 2
U3 → [1+½][1+⅓][1+¼] = (3/2)(4/3)(5/4) = 2,5
Un' → [1+½][1+⅓][1+¼]...[1+1/(n'+1)] = 11
...
Membentuk barisan aritmatika
1,5 , 2 , 2,5 , 3 , ... , 11

Un' = a+(n'-1)b
11 = 1,5+(n'-1)(0,5)
11 = 1,5+0,5n'-0,5
10 = 0,5n'
20 = n'

Pada suku ke 20
U20 → [1+½][1+⅓][1+¼]...[1+1/21] = 11

Maka, nilai n adalah 21.
Semoga membantu..
Maaf jika salah..