Q.Find $\displaystyle\sum_{k=1}^{12} k^3$, and verify that it equals $\left(\displaystyle\sum_{k=1}^{12}k\right)^2$.

[!TLDR] Apply $\sum k^3=\left[\dfrac{n(n+1)}{2}\right]^2$ with $n=12$, and compare with $\left(\sum k\right)^2$. [!ANSWER] $\s …

$\sum_{k=1}^{12}k=\dfrac{12\times13}{2}=78$. By the identity of Section 9, $\sum_{k=1}^{12}k^3=\left(\sum_{k=1}^{12}k\right)^2=78^2=6084$. The verification is therefore automatic and exact by the identity itself — both quantities are, by construction, the same num …

First find $\sum k=\frac{n(n+1)}{2}$, then square it — this directly gives $\sum k^3$ by the boxed identity, with …

Computing $\sum k^3$ by adding twelve individual cubes by hand (rather than using the closed-form identi …