Induction.provide formula. prove it by induction.

Question

I have done the formula but im not sure how to prove this by induction. Can anyone provide me a solution to this part? (solution was not provided)

$$a_n=\frac{n}{4n+1}$$ is the explicit formula I got.

Answer 1

We that the sum is equal to $\frac{n}{4n+1}$ by induction:

Base case: $\frac{1}{1\times 5}=\frac{1}{4\cdot 1 + 5}$

Inductive step:

We must prove $\frac{1}{1\times 5} + \dots + \frac{1}{(4(n+1)-3)(4(n+1)+1)}=\frac{n+1}{4(n+1)+1}=\frac{n+1}{4n+5}$.

By the inductive hypothesis it is equal to:

$\frac{n}{4n+1}+\frac{1}{(4(n+1)-3)(4(n+1)+1)}=\frac{n}{4n+1}+\frac{1}{(4n+1)(4n+5)}=\frac{n(4n+5)+1}{(4n+1)(4n+5)}=\frac{(n+1)(4n+1)}{(4n+1)(4n+5)}=\frac{n+1}{4n+5}$

We are done.