Determine the sum formula for
$$\frac 1{1\cdot 5}+\frac 1{5\cdot 9}+\frac 1{9\cdot 13}+\cdots +\frac1{(4n-3)\cdot (4n+1)}$$ and prove it by induction.
Let us analyse the four first elements of the series:
$$\frac1{1\cdot5}, \frac1{5\cdot9}, \frac1{9\cdot13}, \frac1{13\cdot17}$$
Clearly we see a pattern between every subsequent fraction, and the identity between the second element of the denominator and the first element of the second fraction's denominator. Hence the very first term, with $n=1$ is
$$\frac{1}{n(n+4)}$$ The second term with $n=5$ $$\frac{1}{n(n+4)}$$
and so on.
But how do I arrange these terms into a sum? Thanks
The sum is $\frac{n}{4n+1}$. Use $\frac{1}{(4n-3)(4n+1)}=\frac{1}{4}(\frac{1}{4n-3}-\frac{1}{4n+1})$.