Q.Find the sum $\sum_{r=1}^{30} r = 1 + 2 + 3 + \cdots + 30$.

Use the standard sum $\sum r = \frac{n(n+1)}{2}$ with $n = 30$.

$= \frac{930}{2} = 465$.

Apply $\displaystyle\sum_{r=1}^{n} r = \dfrac{n(n+1)}{2}$ with $n = 30$:

Independent check (AP sum): $1, 2, \ldots, 30$ is an AP with $a = 1$, $l = 30$, $n = 30$, so $S_{30} = \dfrac{30}{2}(1 + 30) = 15 \times 31 = 465$ — matches.

Pairing: $(1 + 30) + (2 + 29) + \cdots$ makes $15$ pairs each summing to $31$: $15 \times 31 = 465$ — the same figure, showing why the formula is $\frac{n(n+1)}{2}$.

Mis-multiplying $30 \times 31$ as $903$ or $960$ instead of $930$ — compute it as $30 \times 31 = 30\times30 + 30 = 900 + 30 = 930$, then halve.