Multiple part problem concerning the proof that $\sum_{k=1}^n k^3=\left(\frac{n(n+1)}{2}\right)^2$ by induction

Question

So I'm having trouble with $c,d$ and $e$.

For $c$ so far I have:

Inductive Hypothesis: $(\frac{n(n+1)}{2})^2 = (\frac{(k+1)(k+2)}{2})^2$ is that correct?

Answer 1

For $n\geq 1$, let $P(n)$ denote the following statement: $$ P(n) : \sum_{k=1}^n k^3=\left(\frac{n(n+1)}{2}\right)^2. $$

(a): $P(1)$ is the statement that $\sum_{k=1}^1 k^3 = \left(\frac{1(1+1)}{2}\right)^2$.

(b): $P(1)$ is true because $\sum_{k=1}^1 k^3 = 1 = \left(\frac{1(1+1)}{2}\right)^2$.

(c): Fix some $\ell\geq 1$ and assume that $P(\ell)$ is true where $$ P(\ell) : \sum_{k=1}^\ell k^3=\left(\frac{\ell(\ell+1)}{2}\right)^2. $$ That statement $P(\ell)$ is the inductive hypothesis.

(d): To be shown is that $P(\ell+1)$ follows where $$ P(\ell+1) : \sum_{k=1}^{\ell+1} k^3=\left(\frac{(\ell+1)(\ell+2)}{2}\right)^2. $$ Starting with the left-hand side of $P(\ell+1)$, \begin{align} \sum_{k=1}^{\ell+1}k^3 &= \color{red}{\sum_{k=1}^\ell k^3}+(\ell+1)^3\tag{by defn. of $\Sigma$}\\[1em] &= \color{red}{\left(\frac{\ell(\ell+1)}{2}\right)^2}+(\ell+1)^3\tag{by $P(\ell)$, the ind. hyp}\\[1em] &= \frac{(\ell+1)^2}{4}[\ell^2+4(\ell+1)]\tag{factor out $(\ell+1)^2/4$}\\[1em] &= \frac{(\ell+1)^2}{4}[(\ell+2)(\ell+2)]\tag{factor quad. polynomial}\\[1em] &= \frac{(\ell+1)^2(\ell+2)^2}{4}\tag{multiply and rearrange}\\[1em] &= \left(\frac{(\ell+1)(\ell+2)}{2}\right)^2, \end{align} we end up at the right-hand side of $P(\ell+1)$.

(e): Steps (a)-(d) show that the formula is true for all positive $n$ because we showed that $P(1)$ was true and that $P(\ell)\to P(\ell+1)$. This is what all we must show for mathematical induction to be a valid method of proof. We have shown this; thus, $P(n)$ must be true for all positive $n$.

Answer 2

No. The inductive hypothesis is $\sum\limits_{k=1}^n k^3 = {({{\frac{{n(n+1)}}{2}}})}^2 $, which we assume is true.

From there: $$ \sum\limits_{k=1}^n k^3 +(n+1) = {({\frac{n(n+1)}{2}})}^2+(n+1) $$ Edit - the above has a typo, it should be: $$ \sum\limits_{k=1}^n k^3 +(n+1)^3 = \left ({\frac{n(n+1)}{2}}\right )^2+(n+1)^3 $$

Try to see if you can go on from here.