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
 A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
a_{n} & \equiv \sum_{k = 1}^{n}{1 \over \pars{4k - 3}\pars{4k + 1}} =
\sum_{k = 1}^{n}\int_{0}^{1}x^{4k - 4}\,\dd x\int_{0}^{1}y^{4k}\,\dd y =
\int_{0}^{1}\int_{0}^{1}\sum_{k = 1}^{n}\bracks{\pars{xy}^{4}}^{k}
\,{\dd x \over x^{4}}\,\dd y
\\[5mm] & =
\int_{0}^{1}y^{3}\int_{0}^{1}\pars{xy}^{4}\,{\pars{xy}^{4n} - 1 \over \pars{xy}^{4} - 1}\,{y\,\dd x \over \pars{xy}^{4}}\,\dd y =
\int_{0}^{1}y^{3}\int_{0}^{y}{x^{4n} - 1 \over x^{4} - 1}\,\dd x\,\dd y
\\[5mm] & =
\int_{0}^{1}{x^{4n} - 1 \over x^{4} - 1}\int_{x}^{1}y^{3}\,\dd y\,\dd x =
\int_{0}^{1}{x^{4n} - 1 \over x^{4} - 1}{1 - x^{4} \over 4}\,\dd x
\\[5mm] & =
{1 \over 4}
\int_{0}^{1}\pars{1 - x^{4n}}\,\dd x = {1 \over 4}\pars{1 - {1 \over 4n + 1}} =
\bbx{n \over 4n + 1}
\end{align}

'Induction' was
  already shown by $\texttt{@Simply Beautiful Art}$.

