If $a_{n+1}=a_{n}^2+3a_n+1$ and $a_1=\frac13$ then find the value of $$\frac{1}{a_1+2}+\frac{1}{a_2+2}+\frac{1}{a_3+2}+\cdots+\frac{1}{a_{11}+2}+\frac{1}{a_{12}+1}$$
I can easily do this by calculating each $a_i$ separately then doing the calculation but that will be a tiresome task.
Is there any other nicer way$?$
Any help is greatly appreciated.
Start from the back. Use the recursion to get,
$$\frac{1}{a_{i+1}+1}=\frac{1}{(a_{i}+1)(a_{i}+2)}.$$ and so,$$\frac{1}{a_{i}+2}+\frac{1}{a_{i+1}+1}=\frac{1}{a_{i}+2}+\frac{1}{(a_{i}+1)(a_{i}+2)}=\frac{1}{(a_{i}+2)}(1+\frac{1}{a_{i}+1})=\frac{a_{i}+2}{(a_{i}+1)(a_{i}+2)}=\frac{1}{a_{i}+1}.$$
Use this for $i=11$ to reduce the sum to
$$\frac{1}{a_{1}+2}+\dots+\frac{1}{a_{10}+2}+\frac{1}{a_{11}+1}.$$
Can you take it from here?