Vtyjkyuk

In Tom Apostol's Calculus Vol 1 book, Apostol presents the following equation:
$$3[1^2+2^2+...+(n-1)^2]+ 3[1 + 2+ . . . + (n - 1)] + (n - 1)=n^3-1^3$$

He then states that the second sum on the left, which is $[1 + 2+ . . . + (n - 1)]$, is a sum of terms in an arithmetic progression, so that part will simplify to $fracn(n-1)2$, thus the final equation becomes this:
$$1^2+2^2+...+(n-1)^2=fracn^33-fracn^22+fracn6$$
What I cannot understand is how Apostol simplified $[1 + 2+ . . . + (n - 1)]$ to $fracn(n-1)2$. As far as I understand, a sum of terms of an arithmetic progression has the following format: $fracn(a+b)2$, with $a$ and $b$ representing the initial and $b$th term of the sequence, respectively. Where does he get $n-1$? Furthermore, how does he simplify the final formula to $1^2+2^2+...+(n-1)^2=fracn^33-fracn^22+fracn6$?

Note that you have $n-1$ terms, not $n$. Thus the sum will be of the form $frac(n-1)(a+b)2$. Now it makes sense.

After substituting this in the equation you get:

$$3sum_k=1^n-1 k^2 = n^3 - 1 - (n-1) - 3fracn(n-1)2 = n^3 - n - frac3n^22 + frac3n2 = n^3 - frac3n^22 + fracn2$$

$$sum_k=1^n-1 k^2 = fracn^33 - fracn^22 + frac n6$$