[1]: https://i.stack.imgur.com/0MnZo.jpg the original equation
as for the image provided, is there a way to simplify? I have reduced the last two portions (the division) down to $$\frac{6}{a+2}$$
So the final equation is the following: $$\frac{2(a+(a+1)+(a+2)+...+2a)}{a^2+3a+2}+\frac{6}{a+2}$$
Can I reduce it even further? the only thing bothering me is the series part. I don't know what to do with that part.
Thanks for any help.
Note $a^2+3a+2=(a+2)(a+1)$, so you might expect the term $(a+1)$ to appear in the numerator after simplifications so cancellation occurs.
We have $$a+(a+1)+(a+2)+...+(2a-2)+(2a-1))+2a$$ $$=a+(a+1)+(a+2)+...+(a+(a-2))+(a+(a-1))+a+a$$ $$=\underbrace{a+a+...+a}_{a+1 \space\space\text{times}}+(1+2+3+...+a-1+a)$$ $$=a(\underbrace{1+1+...+1}_{a+1 \space\space\text{times}})+\frac{1}{2}a(a+1)$$ $$=a(a+1)+\frac{1}{2}a(a+1)=\frac{3}{2}a(a+1)$$
Or note that the $n$-th term is $b_n=a+n$ so we have $$a+(a+1)+(a+2)+...+(2a-2)+(2a-1))+2a$$ $$=\sum_{n=0}^{a}b_n=\sum_{n=0}^{a}a+\sum_{n=0}^{a}n$$ $$=\frac{3}{2}a(a+1).$$
Or as mentioned by @kerf in the comments $$a+(a+1)+(a+2)+...+(2a-2)+(2a-1))+2a$$ $$=(1+2+3+...+a+...+2a)-(1+2+3+...+(a-2)+(a-1))$$ $$=a(2a+1)-\frac{1}{2}a(a-1)$$ $$=\frac{3}{2}a(a+1)$$
