Q.Prove the following by using the principle of mathematical induction for all $n \in N$: $1.3 + 3.5 + 5.7 + \ldots + (2n-1)(2n+1) = \dfrac{n(4n^2+6n-1)}{3}$

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 $\dfrac{(k+1)\big(4(k+1)^2+6(k+1)-1\big)}{3}$.

Adding the next term $(2k+1)(2k+3)=4k^2+8k+3$ to both sides of the hypothesis:

Expand the numerator:

Now expand the target RHS for $n=k+1$: since $4(k+1)^2+6(k+1)-1=4k^2+8k+4+6k+6-1=4k^2+14k+9$, …

Product-term induction requiring full cubic expansion (the factoring trick does not simplify as cleanly here). Add the next term $(2k+1)(2k+3)$, expand both the resulting numerator and the target c …

• Misidentifying the next term: the $(k+1)$-th term is $(2(k+1)-1)(2(k+1)+1)=(2k+1)(2k+3)$, not $(2k-1)(2k+1)$ again.
• Expansion slips in the cubic $k(4k^2+6k-1)=4k^3+6k^2-k$ (sign of the linear term is easy to drop). …