Question

find the value
(-1)^n+(-1)^2n+(-1)^2n+1+(-1)^4n+2 where n is any positive odd integer

Answer 1

$(-1)^{n}+(-1)^{2n}+(-1)^{2n+1}+(-1)^{4n+2}= -1+1-1+1=0$

reason is : 
n is a positive odd integer. So (-1)^n = -1
2n is a positive EVEN integer
$.[(-1)^2]^n = 1^n = 1$
2n+1 is a positive ODD integer 
$(-1)^{2n+1}=[(-1)^2]^n* * (-1)^1}= 1^{n} (-1)=-1$
4n+2 is always a positive even integer.
 $(-1)^{4n+2}=[(-1)^2]^{2n+1}= 1^{2n+1}=1$

Answer 2

Step-by-step explanation:

−1)

n

+(−1)

2n

+(−1)

2n+1

+(−1)

4n+2

=−1+1−1+1=0

reason is :

n is a positive odd integer. So (-1)^n = -1

2n is a positive EVEN integer

.[(-1)^2]^n = 1^n = 1.[(−1)

2

]

n

=1

n

=1

2n+1 is a positive ODD integer

4n+2 is always a positive even integer.

\begin{lgathered}(-1)^{4n+2}=[(-1)^2]^{2n+1}= 1^{2n+1}=1\\\end{lgathered}

(−1)

4n+2

=[(−1)

2

]

2n+1

=1

2n+1

=1