Question

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

Answer 1

-1 raised to the power of any odd number is -1 and that of even number is 1.
Now given that n is odd,therefore,
${ - 1}^{n} + { - 1}^{2n} + { - 1}^{2n + 1} + { - 1}^{2(2n + 1)}$
Now every number multiplied by 2 is an even number, therefore 2n+1 is odd. Therefore the answer is
$- 1 + 1 - 1 + 1 = 0$