Q.Prove the following by using the principle of mathematical induction for all $n \in N$: $1^3 + 2^3 + 3^3 + \ldots + n^3 = \left(\dfrac{n(n+1)}{2}\right)^2$

Let $P(n)$ be the statement

Base case: For $n=1$,

So $P(1)$ is true.

Inductive step: Assume $P(k)$ is true for some $k\ge1$:

We must show $1^3+\ldots+k^3+(k+1)^3=\left(\dfrac{(k+1)(k+2)}{2}\right)^2$.

Adding $(k+1)^3$ to both sides of the hypothesis:

Factor out $(k+1)^2$:

This is exactly $P(k+1)$.

Sum-of-cubes induction. Add $(k+1)^3$ to the squared closed form, factor out the common $(k+1)^2$, and recognize $k^2+4k+4=(k+2)^2$ to reassemble the target closed form.

• Forgetting to square $\frac{k(k+1)}{2}$ correctly, i.e. writing $\frac{k^2(k+1)}{4}$ instead of $\frac{k^2(k+1)^2}{4}$.
• Not spotting that $k^2+4k+4$ factors as the perfect square $(k+2)^2$, leaving the answer in an unsimplified, non-matching form.
• Adding $(k+1)^3$ as $(k+1)^2\cdot(k+1)$ incorrectly when factoring, dropping a factor of $(k+1)$.