+118735 
+118735 This is a really confusing question to start your learning process with.It is probably easier to think of it as 3 seperate questions.
| n | 1 | 2 | 3 | n |
| Tn | \(\frac{2}{5}\) | \(\frac{2}{7}\) | \(\frac{2}{9}\) | |
| Tn | \(\frac{2}{2*\color{red}{1}\color{black}+3}\) | \(\frac{2}{2*\color{red}{2}\color{black}+3}\) | \(\frac{2}{2*\color{red}{3}\color{black}+3}\) | \(\frac{2}{2*\color{red}{n}\color{black}+3}\) |
I already tried to explain this bit. \(T_n=\frac{2}{2\color{red}{n}\color{black}+3}\)
Second one
| n | 1 | 2 | 3 | n |
| Tn | \(\frac{1}{3}\) | \(\frac{1}{4}\) | \(\frac{1}{5}\) | |
| Tn | \(\frac{1}{\color{red}{1}\color{black}+2}\) | \(\frac{1}{\color{red}{2}\color{black}+2}\) | \(\frac{1}{\color{red}{3}\color{black}+2}\) | \(\frac{1}{\color{red}{n}\color{black}+2}\) |
Third one
| n | 1 | 2 | 3 | n |
| Tn | \(\frac{1}{15}\) | \(\frac{1}{28}\) | \(\frac{1}{45}\) | |
| Tn | \(\frac{1}{5*3}\) | \(\frac{1}{7*4}\) | \(\frac{1}{9*5}\) | |
| Tn |
I already worked out how to get the those numbers from the first and second ones.
See if you can makes sense of what I have added here.
Like I said, you probably need to do a whole lot of easier ones to get the idea.
+118735 Practice will make you much better at these.
| n | 1 | 2 | 3 |
| Tn | \(\frac{2}{5}\) | \(\frac{2}{7}\) | \(\frac{2}{9}\) |
notice the pattern 5,7,9
n is only going up by 1 but the denominator of Tn is going up by 2
So a part of the denominor will be 2n
2n+3 works and the top is just 2. So \(T_n=\frac{2}{2n+3} \)
Now lets look at the next bit
\(T_n=\frac{1}{n+2}+\frac{1}{(n+2)(2n+3)}\\ \text{or if you prefer} \\ T_n=\frac{1}{n+2}+\frac{1}{2n^2+7n+6}\\\)
so
\(T_n=\frac{2}{2n+3} =\frac{1}{n+2}+\frac{1}{(n+2)(2n+3)}\\\)
.
