Question

2^(n+2) × 2^(n-1)/2^n(n-1) ÷ 4^n (solve)

Answer 1

2^(-n^2+n+1)

Explanation:

2^(n+2)*2^(n-1)*2^(-n(n-1))=2^((n+2)+(n-1)-(n^2-n))=2^(-n^2+3n+1)

I can do it because they have the same base, so I can add the exponents.

I can write 4^n=2^2n. So I have 2^(-n^2+3n+1) / 2^2n and they have the same base again (2), so we can subtract the exponents.

2^((-n^2+3n+1)-2n)=2^(-n^2+n+1)