Q.Prove the following by using the principle of mathematical induction for all $n \in N$: $\dfrac{1}{2.5} + \dfrac{1}{5.8} + \dfrac{1}{8.11} + \ldots + \dfrac{1}{(3n-1)(3n+2)} = \dfrac{n}{(6n+4)}$

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 $\dfrac{k+1}{6(k+1)+4}=\dfrac{k+1}{2(3k+5)}$.

Adding the next term $\dfrac{1}{(3k+2)(3k+5)}$ to both sides of the hypothesis:

$\frac{k}{2(3k+2)}+\frac{1}{(3k+2)(3k+5)}=\frac{1}{3k+2}\left[\frac{k}{2}+\frac{1}{3k+5}\right]=\frac{1}{3k+2}\cdot\frac{k(3k+5)+2}{2(3k+5)}$ …

Telescoping induction. Rewrite $6k+4=2(3k+2)$ to spot the common factor, add the next term $\frac{1}{(3k+2)(3k+5)}$, factor $(3k+2)$ …

• Not rewriting $6k+4$ as $2(3k+2)$ first, making the common factor with the next term's denominator hard to see.
• Missing the factorization $3k^2+5k+2=(3k+2)(k+1)$ and instead leaving an unsimplified quadratic that doesn't visibly cancel. …