Prove by induction that for all $n \in \Bbb N$ $$\sum_{k\mathop=1}^n k = \frac{n(n+1)}2 \quad \text{and} \quad \sum_{k\mathop=1}^n k^3 = \left(\sum_{k\mathop=1}^n k\right)^2$$
I have proved the first part but can anyone show me how to solve the second part?
Thanks.
We need to prove that $$1^3+2^3+...+n^3=\frac{n^2(n+1)^2}{4}.$$ For $n=1$ it's obvious and for the proof of $$1^3+2^3+...+n^3+(n+1)^3=\frac{(n+1)^2(n+2)^2}{4}$$ it's enough to prove that $$\frac{n^2(n+1)^2}{4}+(n+1)^3=\frac{(n+1)^2(n+2)^2}{4},$$ which is obvious.
Theorem. $$ \sum_{k = 1}^n k^3 = \frac{n^2(n+1)^2}{4} $$
Proof: (by Induction). Let $P_n$ be the proposition: $$P_n: \qquad \sum_{k = 1}^n k^3 = \frac{n^2(n+1)^2}{4} $$
Base Case: $$ \sum_{k = 1}^1 k^3 = 1^3 = 1 $$ $$ \frac{1^2(1+1)^2}{4} = \frac{1\cdot4}{4} = 1 $$ Therefore $P_1$ holds.
Inductive Step: Assume, for the sake of induction, $P_m$, for some arbitrary $m$. Then: \begin{align*} \sum_{k = 1}^m k^3 &= \frac{m^2(m+1)^2}{4} \\ \sum_{k = 1}^m k^3 + (m+1)^3 &= \frac{m^2(m+1)^2}{4} + (m+1)^3 \\ \sum_{k = 1}^{m+1} k^3 &= (m+1)^2 \left[ \frac{m^2}{4} + (m+1) \right] \\ &= (m+1)^2 \cdot \frac{m^2 + 4m + 4}{4} \\ &= \frac{(m+1)^2(m+2)^2}{4} \\ \end{align*} Therefore, $P_m \implies P_{m+1}$.
Conclusion: We have shown: $$ P_1 \wedge (P_m \implies P_{m+1})$$ By mathematical induction, we have that $P_n$ holds for all $n \in \mathbb{N}$.
