# Finding explicit formula of this recursion formula

## Homework Statement

Write an explicit formula for the sequence determined by the following recursion formula.

t$_{1}$= 0; t$_{n}$ = t$_{n-1}$ + $\frac{2}{n(n+1)}$

## The Attempt at a Solution

t$_{1}$ = 0

t$_{2}$ = t$_{1}$ + $\frac{2}{2(2+1)}$
t$_{2}$ = $\frac{1}{3}$

t$_{3}$ = t$_{2}$ + $\frac{2}{3(3+1)}$
t$_{3}$ = $\frac{1}{3}$ + $\frac{2}{3(3+1)}$
t$_{3}$ = $\frac{4}{12}$ + $\frac{2}{12)}$
t$_{3}$ = $\frac{1}{2}$

t$_{4}$ = t$_{3}$ + $\frac{2}{4(4+1)}$
t$_{4}$ = $\frac{1}{2}$ + $\frac{2}{20}$
t$_{4}$ = $\frac{3}{5}$

My sequence is 0, $\frac{1}{3}$, $\frac{1}{2}$, $\frac{3}{5}$ $\cdots$

How do I make an explicit formula if there is no common difference nor a common ratio?

Curious3141
Homework Helper

## Homework Statement

Write an explicit formula for the sequence determined by the following recursion formula.

t$_{1}$= 0; t$_{n}$ = t$_{n-1}$ + $\frac{2}{n(n+1)}$

## The Attempt at a Solution

t$_{1}$ = 0

t$_{2}$ = t$_{1}$ + $\frac{2}{2(2+1)}$
t$_{2}$ = $\frac{1}{3}$

t$_{3}$ = t$_{2}$ + $\frac{2}{3(3+1)}$
t$_{3}$ = $\frac{1}{3}$ + $\frac{2}{3(3+1)}$
t$_{3}$ = $\frac{4}{12}$ + $\frac{2}{12)}$
t$_{3}$ = $\frac{1}{2}$

t$_{4}$ = t$_{3}$ + $\frac{2}{4(4+1)}$
t$_{4}$ = $\frac{1}{2}$ + $\frac{2}{20}$
t$_{4}$ = $\frac{3}{5}$

My sequence is 0, $\frac{1}{3}$, $\frac{1}{2}$, $\frac{3}{5}$ $\cdots$

How do I make an explicit formula if there is no common difference nor a common ratio?

Do a partial fraction decomposition on ##\frac{2}{n(n+1)}##. Let that be ##\frac{A}{n} + \frac{B}{n+1}## (you determine A and B).

Now ##t_n = t_{n-1} + \frac{A}{n} + \frac{B}{n+1}## and ##t_{n-1} = t_{n-2} + \frac{A}{n-1} + \frac{B}{n}##.

Substitute the latter expression into the first and see what happens. Now continue successive substitution until you arrive at ##t_1##.

haruspex
is that calculus? :$haruspex Science Advisor Homework Helper Gold Member 2020 Award is that calculus? :$