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

Compute both sides for $n=6$ independently and compare.

Left-hand side (direct addition): $\sum_{i=1}^{6} i = 1+2+3+4+5+6$. Adding in order: $1+2=3$, $+3=6$, $+4=10$, $+5=15$, $+6=21$. So the direct sum is $21$.

Right-hand side (formula): $\dfrac{n(n+1)}{2}$ at $n=6$ gives $\dfrac{6\times7}{2}=\dfrac{42}{2}=21$. …

Alternative — pairing check: pair $1+6=7$, $2+5=7$, $3+4=7$ — three pairs each summing to $7$, giving $3\times7=2 …

A common slip is substituting $n=6$ into the formula as $\dfrac{6(6+1)}{2}$ but then mis-multiplying $6\times7$ as $36$ instead of $42$ — al …