Q.Prove the following by using the principle of mathematical induction for all $n \in N$: $\dfrac{1}{1.4} + \dfrac{1}{4.7} + \dfrac{1}{7.10} + \ldots + \dfrac{1}{(3n-2)(3n+1)} = \dfrac{n}{(3n+1)}$

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 $\dfrac{k+1}{3k+4}$.

Adding the next term $\dfrac{1}{(3k+1)(3k+4)}$ to both sides of the hypothesis:

$\frac{k}{3k+1}+\frac{1}{(3k+1)(3k+4)}=\frac{1}{3k+1}\left[k+\frac{1}{3k+4}\right]=\frac{1}{3k+1}\cdot\frac{k(3k+4)+1}{3k+4}$ …

Telescoping induction. Add the next term $\frac{1}{(3k+1)(3k+4)}$, factor $(3k+1)$ out of the resulting numerator $3k^2+4k+1=(3k+1)(k+1)$ …

• Not spotting the factorization $3k^2+4k+1=(3k+1)(k+1)$ and leaving the fraction unsimplified.
• Misreading the denominator pattern $(3n-2)(3n+1)$, using $3n-1$ instead of $3n-2$ for the next term. …