Answer:
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)
