If a^n ≠ 0 and n is a positive integer, is n odd?a^n ≠ 0 and n is a positive integer implies that a ≠ 0.
(1) a^n + a^(n+1) < 0 --> \((a+1)a^n<0\). Two cases:
\(a+1>0\) and \(a^n<0\). From first inequality we have that \(a>-1\). If a is a negative number (say -1/2), then \((negative)^n\) to be negative n must be odd.
\(a+1<0\) and \(a^n>0\). From first inequality we have that \(a<-1\), so a is a negative number. Now, \((negative)^n\) to be positive n must be even.
Not sufficient.
(2) a is an integer. Clearly insufficient.
(1)+(2) Consider the same two cases but now take into account that a is an integer:
\(a+1>0\) and \(a^n<0\). From first inequality we have that \(a>-1\). Since we know that \(a\neq{0}\), then a must be a positive integer (1, 2, 3, ...). Next, \((positive)^n\) cannot be negative, thus this case is out.
\(a+1<0\) and \(a^n>0\). From first inequality we have that \(a<-1\), so a is a negative integer. Now, \((negative)^n\) to be positive n must be even.
Sufficient.
Answer: C.
Hope it's clear.