Prove, using mathematical induction, that for all $n ≥ 1$, the sum of the cubes of the first $n$ natural numbers is equal to $(n(n+1)/2)^2$.

Steps to Solve

1. Base Case: Verify for $n=1$

Show that the formula holds true when $n=1$. Left-hand side (LHS): $1^3 = 1$ Right-hand side (RHS): $\left(\frac{1(1+1)}{2}\right)^2 = \left(\frac{2}{2}\right)^2 = 1^2 = 1$ Since LHS = RHS, the formula holds true for $n=1$.

2. Inductive Hypothesis: Assume true for $n=k$

Assume that the formula is true for some arbitrary positive integer $k$. This means we assume: $$1^3 + 2^3 + \dots + k^3 = \left(\frac{k(k+1)}{2}\right)^2$$ This is our inductive hypothesis.

3. Inductive Step: Prove true for $n=k+1$

We want to prove that the formula holds true for $n=k+1$. That is we need to show: $$1^3 + 2^3 + \dots + k^3 + (k+1)^3 = \left(\frac{(k+1)(k+1+1)}{2}\right)^2 = \left(\frac{(k+1)(k+2)}{2}\right)^2$$ Starting with the left-hand side and using the inductive hypothesis: $$1^3 + 2^3 + \dots + k^3 + (k+1)^3 = \left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3$$ Now, we manipulate the expression to match the right-hand side: $$ \left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3 = \frac{k^2(k+1)^2}{4} + (k+1)^3 = \frac{k^2(k+1)^2 + 4(k+1)^3}{4}$$ Factor out $(k+1)^2$ from the numerator: $$ \frac{(k+1)^2[k^2 + 4(k+1)]}{4} = \frac{(k+1)^2(k^2 + 4k + 4)}{4} = \frac{(k+1)^2(k+2)^2}{4}$$ Rewrite the expression: $$ \frac{(k+1)^2(k+2)^2}{4} = \left(\frac{(k+1)(k+2)}{2}\right)^2$$ This is exactly what we wanted to show for $n=k+1$.

4. Conclusion

Since the base case holds, and assuming the formula holds for $n=k$, we have shown that it also holds for $n=k+1$, by the principle of mathematical induction, the formula is true for all integers $n \ge 1$.