Q.Evaluate $\sum_{r=1}^{12} r^2 = 1^2 + 2^2 + \cdots + 12^2$.

Use $\sum r^2 = \frac{n(n+1)(2n+1)}{6}$ with $n = 12$.

$= \frac{3900}{6} = 650$.

Apply $\displaystyle\sum_{r=1}^{n} r^2 = \dfrac{n(n+1)(2n+1)}{6}$ with $n = 12$ (so $2n + 1 = 25$):

Simplify by cancelling the $6$ into the $12$: $\dfrac{12}{6} = 2$, so the expression becomes $2 \times 13 \times 25 = 2 \times 325 = 650$.

Independent check (full multiply then divide): $12 \times 13 = 156$; $156 \times 25 = 3900$; $3900 \div 6 = 650$ — matches.

Using $(2n - 1) = 23$ instead of $(2n + 1) = 25$ — the factor is $2n + 1$, computed by doubling $n$ and then adding one. The wrong factor would give $\frac{12 \times 13 \times 23}{6} = 598$.