Prove by mathematical induction
\[
\frac{1}{1 \times 5} + \frac{1}{5 \times 9} + \cdots + \frac{1}{(4n-3)(4n+1)}
= \frac{n}{4n+1}
\]
for all natural numbers n > 0.
Prove the statement is true for a particular n. Here n = 1 looks a reasonable choice.
To show this is true for n=1, observe that the left hand side = 1/5 and the right hand side = 1/(4.1+1) = 1/5.
So the statement is true for n=1.
Assume the statement is true for n = k and show it is true for n = k + 1.
To show the statement is true for n = k+1, call the LHS \( S_{k+1} \). Then
\[ \begin{align}
S_{k+1}
&= \frac{1}{1 \times 5} + \frac{1}{5 \times 9} + \cdots + \frac{1}{(4k+1)(4k+5)} \\
\end{align} \ \]
Use the inductive hypothesis to simplify the right hand side.
By the inductive hypothesis,
\[ \begin{align}
S_{k+1} &= S_k + \frac{1}{(4k+1)(4k+5)} \\
&= \frac{k}{4k+1} + \frac{1}{(4k+1)(4k+5)} \\
&= \frac{k(4k+5)+1}{(4k+1)(4k+5)} \\
&= \frac{4k^2+5k+1}{(4k+1)(4k+5)} \\
&= \frac{(4k+1)(k+1)}{(4k+1)(4k+5)} \\
&= \frac{k+1}{4(k+1) + 1} \\
\end{align} \ \]
Now the RHS has the same form as the RHS of the original expression with k+1 substituted for n. So the statement is true for n = k+1.
By the principle of mathematical induction,
The statement is true for \( n \ge 1 \).