Question

Soal b, sederhanakan saya bingung bagian -2 nya...

Answer 1

= (2^[n+2])²/(2ⁿ×2^[n+2]) - (2² × 2²ⁿ)/(2ⁿ × 2^[n+2])
= (2^[2n+4])/(2^(2n+2) - (2^[2n+2])/(2^[2n+2])
= 2^(2n+4-2n-2) - 1
= 2² - 1
= 4 - 1
= 3