Q.Verify that $\sum_{i=1}^{5} i^2 = \dfrac{n(n+1)(2n+1)}{6}$ holds for $n=5$.

Compute the direct sum of squares and the formula's value for $n=5$, and compare.

Left-hand side (direct addition of squares): $\sum_{i=1}^{5} i^2 = 1^2+2^2+3^2+4^2+5^2=1+4+9+16+25$. Adding: $1+4=5$, $+9=14$, $+16=30$, $+25=55$. So the direct sum is $55$. …

Alternative computation order — multiply the numerator fully first: $5\times6\times11=330$, then divide by $6$: $330/6=55$ — the same result reached without cancel …

A frequent error is using $(2n-1)$ instead of $(2n+1)$ (giving $9$ instead of $11$ at $n=5$), which would produce a wrong right-hand side of $45$ instead of $ …