Q.Prove the following by using the principle of mathematical induction for all $n \in N$: $\dfrac{1}{3.5} + \dfrac{1}{5.7} + \dfrac{1}{7.9} + \ldots + \dfrac{1}{(2n+1)(2n+3)} = \dfrac{n}{3(2n+3)}$

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

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

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

Telescoping induction. Add the next term $\frac{1}{(2k+3)(2k+5)}$, factor $(2k+3)$ out of the nume …

• Confusing this problem's denominators $(2n+1)(2n+3)$ with the similar-looking Exercise 16 pattern $(3n-2)(3n+1)$ and mixing up the algebra. …