Question

is \[(-1)^{1-n}=(-1)^{n+1}\]

Answer 1

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

Answer 2

which equality are you wanting?
are there any restrictions on n?

Answer 3

sorry
n is intergers;

Answer 4

Plug in values for n and see... If n =0 you get what? -1 for both?

Answer 5

Just plug in n values.

Answer 6

Let 1 - n be a

\[(-1)^a = (-1)^{a+2n}\]
If a is odd,
then a + 2n is odd since odd + even = odd.
If a is even,
then a + 2n is even since even + even = even.

Answer 7

i'm still looking at \[(-1)^{1-n}=(-1)^{n+1}\]

Answer 8

so for n={1,2,3,...}

\[(-1)^{n+1} = (-1)^{n-1} = (-1)^{1-n} = (-1)^{1+n}\]
?

Answer 9

lol i got confused

Answer 10

just plug in values like male said to see what happens with whatever you are checking is true

Answer 11

If n is odd, then
n + 1 is even
n -1 is also even
1 - n is even since odd - odd = even

Answer 12

money posted some good stuff too

Answer 13

well i plugged in values and that told me the expressions are all equivalent , i was just making sure that i can substitute these terms at will.

Answer 14

did you look at what money wrote?

Answer 15

does anything change if
if the \[(-1)^{n+1}\]is in a summation over n \[ \sum_{n=1}^∞ \]

Answer 16

\[\text{ if n is odd then n+1 and n-1 and 1-n are even}\]
\[(-1)^{n+1}=(-1)^{n-1}=(-1)^{1-n}\]
all these powers are even integers
which means we have 1=1=1

so what if n is even then n+1 and n-1 and 1-n are odd
\[(-1)^{n-1}=(-1)^{1-n}=(-1)^{n+1}\]
since -1=-1=-1

Answer 17

im guessing it doesnt change anything

Answer 18

thank you all for your patients