Q.Prove the following by using the principle of mathematical induction for all $n \in N$: $1^2 + 3^2 + 5^2 + \ldots + (2n-1)^2 = \dfrac{n(2n-1)(2n+1)}{3}$

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 the sum up to $(k+1)$ equals $\dfrac{(k+1)(2k+1)(2k+3)}{3}$.

Adding the next term $(2k+1)^2$ to both sides of the hypothesis:

Expand and factor the bracket: …

Sum-of-odd-squares induction. Add the next odd square $(2k+1)^2$, factor out the common $(2k+1)$, and factor the resulting quadratic $2 …

• Misidentifying the next term: after $(2k-1)^2$ the next odd number squared is $(2k+1)^2$, not $(2k+3)^2$.
• Sign error expanding $k(2k-1)=2k^2-k$ (dropping the minus sign on $k$). …