The $n{th}$ of a sequence is given by $T_n =\frac{3}{4n^2+8n+3}.$

Question

The $n{th}$ of a sequence is given by $T_n =\frac{3}{4n^2+8n+3}. $ How to find an expression for the sum of the first $n$ terms of the sequence?

Answer 1

Use the fact that$$T_n=\frac32\left(\frac1{2n+1}-\frac1{2(n+1)+1}\right).$$

Answer 2

$$T_n =\frac{3}{4n^2+8n+3} = \frac{3}{(2n+1)(2n+3)}$$

$$\begin{align}\sum_{k=1}^n \frac{3}{(2k+1)(2k+3)} &= \frac32.\left(\frac {1}{2k+1} - \frac{1}{2k+3}\right)\\ & = \frac32\left(\frac 13 - \frac {1}{2n+3}\right)\\ & = \boxed{\color {red}{\frac{n}{2n+3}}}\end{align}$$