<s>[INST] How find this $a_{n}$ closed form $a_{n}=\frac{1}{1+b}a_{n-1}+\frac{b}{1+b^{n+1}},n\ge 2$ Question:
let $b\neq 0,1,-1$,and sequence $\{a_{n}\}$,such
$$a_{1}=\dfrac{1}{(1+b)^2}$$
and such
$$a_{n}=\dfrac{1}{1+b}a_{n-1}+\dfrac{b}{1+b^{n+1}},n\ge 2$$
Find the $a_{n}$ closed form?
My try: since
$$x=\dfrac{1}{1+b}x+\dfrac{b}{1+b^{n+1}}$$
then
$$x=\dfrac{1+b}{1+b^{n+1}}$$
then 
$$a_{n}-\dfrac{b+1}{1+b^{n+1}}=\dfrac{1}{1+b}a_{n-1}-\dfrac{1}{1+(1+b)^{n+1}}=\dfrac{1}{1+b}\left(a_{n-1}-\dfrac{b+1}{b^{n+1}+1}\right)$$
then I can't 
because this right hand is not 
$$\dfrac{1}{1+b}\left(a_{n-1}-\dfrac{b+1}{b^{n}+1}\right)$$
so if let
$$b_{n}=a_{n}-\dfrac{b+1}{1+b^{n+1}}$$
But the right is not $\dfrac{1}{b+1}b_{n-1}$ 
so I can't,Thank you for you help!
 [/INST] Let $$\dfrac{1}{1+b}=k$$
And let
$$f(x) = \dfrac{b}{1+b^{x+1}}$$
Then, $$a_{n}=ka_{n-1}+\dfrac{b}{1+b^{n+1}}$$
$$a_{n}-ka_{n-1}=\dfrac{b}{1+b^{n+1}}$$
$$\sum\limits_{x=2}^n (\dfrac{1}{k^x}(a_x+ka_{x-1})) = \sum\limits_{x=2}^n \dfrac{1}{k^x}f(x)$$
$$\sum\limits_{x=2}^n (\dfrac{a_x}{k^x} - \dfrac{a_{x-1}}{k^{x-1}}) = \sum\limits_{x=2}^n f(x)$$
$$(\sum\limits_{x=2}^{n-1}\dfrac{a_x}{k^x} - \sum\limits_{x=2}^{n-1}\dfrac{a_x}{k^x}) + (\dfrac{a_{n}}{k^x} - \dfrac{a_{1}}{k^1}) = \sum\limits_{x=2}^n \dfrac{1}{k^x}f(x)$$
$$\dfrac{a_{n}}{k^n} = \dfrac{a_{1}}{k^1} + \sum\limits_{x=2}^n \dfrac{1}{k^x}f(x)$$
$$a_{n} = \dfrac{1}{(1+b)^{n + 1}} + \dfrac{1}{(1+b)^{n}}\sum\limits_{x=2}^n \dfrac{1}{k^x}f(x)$$
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