<s>[INST] Induction: $\frac1{1 \cdot 5} + \frac1{5 \cdot 9} + \cdots + \frac1{(4n-3)(4n+1)}$ 
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.

Hi, I'm trying to answer this question. I was not provided with a solution.
 [/INST] $$a_n=\frac{1}{4}\sum_{k=1}^n\left(\frac{1}{4k-3}-\frac{1}{4k+1}\right)=$$
$$=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{4n-3}-\frac{1}{4n+1}\right)=$$
$$=\frac{1}{4}\left(1-\frac{1}{4n+1}\right)=\frac{n}{4n+1}$$
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