2^ (n+2) * 2^ (n-1) / 2^n (n-1) : 4^n (solve)

Question

Mathematics

Milo Vaughan

29 August, 23:30

2^ (n+2) * 2^ (n-1) / 2^n (n-1) : 4^n (solve)

+4

Answers (1)

Know the Answer?

Copper · Answer 1

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

Step-by-step 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)