Consider the following polynomial: \[ P(x) = \prod_{i=1}^n (ix^i + (i+1)x^{i+1}) = x^{\frac{n(n+1)}{2}}\cdot (1+2x) \cdot (2+3x) \dots \cdot (n+(n+1)x). \]Each individual term corresponds to an interesting sequence (its coefficient is the product of the members, and its exponent is the sum of the members). Therefore, Alice's sum, $A$, equals the sum of the coefficient of odd-exponent $x$, while Bob's sum, $B$, equals the sum of the coefficient of even-exponent $x$. Therefore: \[A = \frac{P(1)-P(-1)}{2}, B = \frac{P(1)+P(-1)}{2}.\]Therefore, it all depends on $P(-1)$. We see that: $P(-1) = (-1)^{\frac{n(n+1)}{2}} \times (-1)^n = (-1)^{\frac{n(n+3)}{2}}$. If $n \equiv 0,1 \pmod 4$, $n(n+3)/2$ is even, otherwise it's odd. Therefore, we can conclude that: If $n \equiv 0, 1 \pmod 4$, Bob's number is greater by $1$, If $n \equiv 2, 3 \pmod 4$, Alice's number is greater by $1$.