Question

is (-2)^(n-1) the same as 2^(n-1). If not can we simplify it to make it the same

Answer 1

It depends on what n-1 evaluates to. If n-1 is an odd number (-2)^(n-1) will be negative. However, all values of 2^(n-) will be positive

Answer 2

for example if i have this summation\[\sum_{0}^{\infty} \frac{ n*(-2)^{n-1} }{ 2*2^{n-1} }\]

Answer 3

can i take out the negative from the numerator

Answer 4

You could do -1(n * (-2)^n-1) and distribute that to remove it. that would make it \[-\frac{ n * 2^{(n-1)} }{ 2* 2^{(n-1)} }\]

Answer 5

\[(-2)^{n-1}=(-1\times2)^{n-1}=(-1)^{n-1}\times2^{n-1}\]

Answer 6

then if i take the limit of that as n goes to infity, i would get negative infiity right

Answer 7

I'm not too sure, just learning limits. Sorry