Q.Prove the following by using the principle of mathematical induction for all $n \in N$: $\dfrac{1}{1.2.3} + \dfrac{1}{2.3.4} + \dfrac{1}{3.4.5} + \ldots + \dfrac{1}{n(n+1)(n+2)} = \dfrac{n(n+3)}{4(n+1)(n+2)}$

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 $\dfrac{(k+1)(k+4)}{4(k+2)(k+3)}$.

Adding the next term $\dfrac{1}{(k+1)(k+2)(k+3)}$ to both sides of the hypothesis:

Expand and factor the numerator:

$k(k+3)^2+4=k(k^2+6k+9)+4=k^3+6k^2+9k+4.$ …

Telescoping induction with a cubic factorization. Add the next term, combine over a common denominator, and factor the resulting cubic $k^3+6k^2+9k+4=(k+1)^2(k+4)$ ( …

• Failing to spot the repeated factor $(k+1)^2$ in the cubic numerator, leaving an unsimplified expression that appears not to match the target.
• Sign/expansion errors expanding $k(k+3)^2$ — must expand $(k+3)^2=k^2+6k+9$ first, then multiply by $k$. …