Question

Using mathematical induction, prove that $\frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + ... + \frac{1}{(2n - 1)(2n + 1)} = \frac{n}{2n + 1}$ for all n ∈ N

Answer 1

Mathematical Induction

We will use the Principle of Mathematical Induction for proving a statement.

Let the given statement be P(n).

$\displaystyle\sf P(n): \frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\dots+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$

Checking P(1):

$\displaystyle\sf P(1):\\\\\\ \begin{array}{l|l}\mathbb{LHS} & \mathbb{RHS}\\\cline{1-2}&\\=\dfrac{1}{1.3}& = \dfrac{1}{2(1)+1}\\\\\ =\dfrac{1}{3}&= \dfrac{1}{3}\end{array}$

Thus, P(1) is true.

Suppose P(k) is true.

$\sf\displaystyle\implies\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\dots +\frac{1}{(2k-1)(2k+1)}=\frac{k}{2k+1} \quad \textsf{------ (1)}$

We now need To Prove for P(k+1):

To Prove:

$\sf\displaystyle\implies\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\dots +\frac{1}{(2(k+1)-1)(2(k+1)+1)}=\frac{k+1}{2(k+1)+1}$

Consider the LHS:

$\sf\displaystyle\mathbb{LHS}\\\\\\ =\underbrace{\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\dots +\frac{1}{(2k-1)(2k+1)}}_{\textsf{Use Result (1)}}+\frac{1}{(2(k+1)-1)(2(k+1)+1)}\\\\\\ = \frac{k}{2k+1}+\frac{1}{(2k+1)(2k+3)}\\\\\\ = \frac{1}{2k+1}.\left(k+\frac{1}{2k+3}\right)\\\\\\ = \frac{1}{2k+1} . \left( \frac{2k^2+3k+1}{2k+3}\right)\\\\\\ =\frac{2k^2+2k+k+1}{(2k+1)(2k+3)}\\\\\\ = \frac{2k(k+1)+1(k+1)}{(2k+1)(2k+3)}\\\\\\ = \frac{(k+1)\cancel{(2k+1)}}{\cancel{(2k+1)}(2k+3)}\\\\\\ = \frac{k+1}{2(k+1)+1}$

$=\mathbb{RHS}$

Thus, P(k+1) is true provided P(k) is true.

Now, P(1) is true. P(k) is true $\implies$ P(k+1) is true.

Hence, P(n) is true for all n $\in \mathbb{N}$

Hence Proved, by Principle of Mathematical Induction.