Find the numbers that would satisfy an upper index, that would lead it to be divisible by 3

Question

\begin{equation}\displaystyle{\sum_\limits{i=0}^n}i\end{equation}

Find the values of n for which the summation is divisible by 3. If Possible write a general formula.

Answer 1

First, this is equivalent to:

$\sum_{i = 1}^{n}i = 1 + 2 + ...+ (n - 1) + n$

Using the formula for the sum of an arithmetic series:

$1 + 2 + ...+(n-1)+n = \frac{n(n + 1)}{2}$

Obviously this is an integer, because one of $n$ and $n+1$ must be even (divisible by $2$).

Now, we need $\frac{n(n+1)}{2}$ to be divisible by $3$. This is true if and only if one of $n$ and $n + 1$ is divisible by 3. In other words, $n$ must be a multiple of $3$ or one less than a multiple of $3$. Using modular expressions for compactness, the solution set is

$\boxed{\big\{n : n\equiv 0\pmod{3} \text{ or }n\equiv 2\pmod{3}\big\}}$

Answer 2

\begin{align*} \begin{cases} S = 1 & + & 2 & + & 3 & + & \ldots & + & n\\\\ S = n & + & (n-1) & + & (n-2) & + & \ldots & + & 1 \end{cases} \Longrightarrow 2S = n(n+1) \Longrightarrow S = \frac{n(n+1)}{2} \end{align*}