Guess the formula for $\sum\frac 1{(4n-3)(4n+1)}$ and prove by induction 
For $n \ge 1$, let
  $$a_n = \dfrac1{1 \cdot 5} + \dfrac1{5 \cdot 9} + \cdots + \dfrac1{(4n-3)(4n+1)}.$$
  Guess a simple explicit formula for $a_n$ and prove it by induction.

So I have guessed the formula as : $$\frac{n}{4n+1}$$ 
But wasn't sure how to prove it by induction

Base case:
$$a_1=\frac1{4(1)+1}\color{green}\checkmark$$
Now the inductive step:
$$\begin{align}a_{n+1}&=a_n+\frac1{(4n+1)(4n+5)}\\&=\frac n{4n+1}+\frac1{(4n+1)(4n+5)}\\&=\frac{n(4n+5)}{(4n+1)(4n+5)}+\frac1{(4n+1)(4n+5)}\\&=\frac{4n^2+5n+1}{(4n+1)(4n+5)}\\&=\frac{(4n+1)(n+1)}{(4n+1)(4n+5)}\\&=\frac{n+1}{4(n+1)+1}\color{green}\checkmark\end{align}$$
and we are done.
