Dimensional analysis of the equation $s_{n}= u+ \frac{1}{2}(2n-1)a$ It is a question in my textbook to see if the equation 
$s_{n}= u+ \frac{1}{2}(2n-1)a$
is dimensionally correct where $s_{n}$ is the distance travelled by a uniformly accelerating body with initial velocity $u$ and acceleration $a$ in the nth second. 
The book says that the unit of $s_{n}$ is that of velocity as it is distance per second. I differ in reasoning because in the derivation of the formula the seconds unit of time is ignored, or, the original formula should be:
$s_{n}= u \cdot 1s+ \frac{1}{2}(2n \cdot 1s-1s^2)a$
The book is released by our government so I have my reasons for doubting my reasoning. Any help would be appreciated.
 A: The book is correct. The confusion is probably because of exactly what the variables in the equation mean.
The "distance traveled in the $n$th second" is the distance between the positions at time $t = n$ and $t = n-1$. Using the standard SUVAT formula $s = s_0 + ut + \frac 1 2 at^2$, that means $$s_n = \left(s_0 + un + \tfrac 1 2 a n^2\right) - \left(s_0 + u(n-1) + \tfrac 1 2 a(n-1)^2\right)$$ which simplifies to the OP's equation.
You have to be careful about the dimensions of the "constants" here. The $\frac 1 2$'s are just numbers, but the $1$'s in the $(n-1)$ terms represent $1$ second, not the dimensionless number $1$.
So when you simplify $un - u(n-1)$ to just $u$, that really means "the velocity $u$ times $1$ second", which is a distance, not a velocity.
It may be more "correct" to write this as $u \cdot 1\text{s}$ instead of just $u$, as the OP proposes, but the book's notation is commonly used, so you need to get used to it. Writing $u \cdot 1\text{s}$ can also cause confusion if someone things that "s" is a variable, not an abbreviation for "seconds". And if they think the "$\cdot$" means a scalar product of two vectors (which the OP might not know about yet) that might cause even more confusion.
