Question

3. Find the value of (-1)^n+(-1)^2n+(-1)^4n+2
where n is any
Positive odd integer.​

Answer 1

Answer:

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

Step-by-step explanation:

$(-1)^n =-1 \because \ n \ is \ odd \ number.$

$(-1)^{2n} =1 \because \ 2n \ is \ even \ number.$

$(-1)^{4n+2} =1 \because \ 4n+2 \ is \ even \ number.$

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

Answer 2

Step-by-step explanation:

if we put the value of n is 1

= -1^1 + -1^(2×1)+ -1^(4×1+2)

= -1 + 1 + 1

= 1