The value of n for which 704+12(704)+14(704)+…up n terms =1984−12(1984)+14(1984)-.... up to n terms is

According to the given condition(704) 1−12n=(1984)(2)31−(−1)n2n(2)⇒        128=21122n−1984(−1)n2nIf n is odd, we get 2n=32 or n=5If n is even, we get 128=128/2n⇒n=0.