Prove by induction that $a_{n-1}=\frac{na_{n}+1}{a_{n}+n}$ has a solution $a_{n+1}=\frac{4a_{1}-n(n+3)(1-a_{1})}{n(n+3)(1-a_{1})+4}$
By induction on $n$. The base case is when $n=1$: $a_0=\frac{a_1+1}{a_1+1}$
We assume that $k$ is a positive integer such that $a_{k-1}=\frac{ka_{k}+1}{a_{n}+n}$
Now, $a_{k}=\frac{(k+1)a_{k+1}+1}{a_{k+1}+(k+1)}$
I have tired to $a_k-a_{k-1}$ but it did not work. Would someone please help me out with that! Thank you !
We have $$a_{n-1}=\frac{na_n+1}{a_n+n}\implies a_{n-1}(a_n+n)=na_n+1\implies a_n(a_{n-1}-n)=1-na_{n-1}$$ Now $$a_{n-1}-n=0\implies 1-n^2=0\implies n=\pm 1$$ So, if $n\ge 2$, then $a_{n-1}-n\not=0$, so $$a_n=\frac{1-na_{n-1}}{a_{n-1}-n}$$
The inductive step :
Suppose that $$a_{n+1}=\frac{(n^2+3n+4)a_1-n(n+3)}{-n(n+3)a_{1}+n^2+3n+4}$$ Then, $$\begin{align}a_{n+2}&=\frac{1-(n+2)a_{n+1}}{a_{n+1}-(n+2)}\\\\&=\frac{1-(n+2)\frac{(n^2+3n+4)a_1-n(n+3)}{-n(n+3)a_{1}+n^2+3n+4}}{\frac{(n^2+3n+4)a_1-n(n+3)}{-n(n+3)a_{1}+n^2+3n+4}-(n+2)}\\\\&=\frac{-n(n+3)a_1+n^2+3n+4-(n+2)((n^2+3n+4)a_1-n(n+3))}{(n^2+3n+4)a_1-n(n+3)-(n+2)(-n(n+3)a_1+n^2+3n+4)}\\\\&=\frac{-(n+1)(n^2+5n+8)a_{1}+(n+1)^2(n+4)}{(n+1)^2(n+4)a_1-(n+1)(n^2+5n+8)}\\\\&=\frac{-(n^2+5n+8)a_{1}+(n+1)(n+4)}{(n+1)(n+4)a_1-(n^2+5n+8)}\\\\&=\frac{(n^2+5n+8)a_{1}-(n+1)(n+4)}{-(n+1)(n+4)a_1+(n^2+5n+8)}\\\\&=\frac{((n+1)^2+3(n+1)+4)a_{1}-(n+1)(n+4)}{-(n+1)(n+4)a_1+(n+1)^2+3(n+1)+4}\end{align}$$