Guess the formula for $\sum\frac 1{(4n-3)(4n+1)}$ and prove 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.

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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 $\texttt{@Simply Beautiful Art}$.