Write the first six terms of the sequence.$f\left(n\right)=\left(-2\right)^{n+1}$f(n)=(−2)n+1​The first six terms of the sequence are , , ,

To find the first six terms of the sequence defined by $f(n)=(-2)^{n+1}$, we substitute the values of n from 1 to 6 into the function:

1. For n=1, $f(1)=(-2)^{1+1}=(-2)^2=4$
2. For n=2, $f(2)=(-2)^{2+1}=(-2)^3=-8$
3. For n=3, $f(3)=(-2)^{3+1}=(-2)^4=16$
4. For n=4, $f(4)=(-2)^{4+1}=(-2)^5=-32$
5. For n=5, $f(5)=(-2)^{5+1}=(-2)^6=64$
6. For n=6, $f(6)=(-2)^{6+1}=(-2)^7=-128$

So, the first six terms of the sequence are 4, -8, 16, -32, 64, -128.