Q.Verify that $\sum_{i=1}^{4} i^3 = \left(\dfrac{n(n+1)}{2}\right)^2$ holds for $n=4$.

Compute the direct sum of cubes and the formula's value for $n=4$, and compare.

Left-hand side (direct addition of cubes): $\sum_{i=1}^{4} i^3 = 1^3+2^3+3^3+4^3=1+8+27+64$. Adding: $1+8=9$, $+27=36$, $+64=100$. So the direct sum is $100$.

Right-hand side (formula): $\left(\dfrac{n(n+1)}{2}\right)^2$ at $n=4$ gives $\left(\dfrac{4\times5}{2}\right)^2=\left(\dfrac{20}{2}\right)^2=10^2=100$.

Both sides equal $100$, so the formula is verified for $n=4$. …

A common slip is squaring $n(n+1)$ first and THEN dividing by $2$ (i.e. computing $\dfrac{(n(n+1))^2}{2}$ instead of $\left(\dfrac{n(n+1)}{2}\right)^2$) — order matters here: divide by $2$ first, th …