Mathematical notation in Discrete Mathematics 
The sum of the first $n$ natural numbers, starting at $1$, is $$\frac{n(n + 1)(2n + 1)}{6}.$$

Would this translate to...
For any $n$ a natural number, starting at $1$, the sum of the first $n$ natural numbers implies the equation?
Is it OK to assume that it's for all $n$? Or would that be wrong?
 A: "Yes", except that the formula given is the sum of the squares of the natural numbers, from 1 to n.
A: Incorrect.   It should be the sum for the squares of the first $n$ non-zero natural numbers equals that.
$${\sum_{k=1}^n k =\dfrac{n(n+1)}{2!}\\\bbox[1ex, border:pink solid 1pt]{\sum_{k=1}^n k^2 =\dfrac{n(n+1)(2n+1)}{3!}}\\\sum_{k=1}^n k^3 =\dfrac{6n^2(n+1)^2}{4!}}$$
A: Where did you read that? But more importantly: have you tried to derive the equation yourself?
The sum of the first two natural numbers is 3. The sum of the first three natural numbers is 6. The sum of the first four natural numbers is 10. The sum of the first five natural numbers is 15. Hmm, that looks like the triangular numbers.
If you look these up in the OEIS, you will verify that $$\binom{n + 1}{2} = \frac{n^2 + n}{2} = \sum_{i = 1}^n i.$$
None of those look like the formula you copied. Compute $$\frac{n(n + 1)(2n + 1)}{6} = \frac{2n^3 + 3n^2 + n}{6}$$ for a few $n$, say $0 < n < 8$ and look that up in the OEIS: the square pyramidal numbers should be the very first result, and indeed we see that $$\frac{n(n + 1)(2n + 1)}{6} = \sum_{i = 1}^n i^2.$$ That superscript 2 makes a big difference in the summation — does that count as a pun?

In both summations above you can substitute $i = 0$ and that won't change the result.
