How to solve the recurrence $a_{n}=\frac{n+1}{n} a_{n-1}+3n+3$?

Question

I tried to solve this recurrence by taking out $n+1$ as a common in the RHS, but still have $n \cdot a_n$ and $a_n$

Answer 1

Hint

$$a_{n}=\frac{n+1}na_{n-1}+3n+3,$$ $$ na_n= (n+1)a_{n-1} + 3n(n+1) $$

$$ \frac{a_n}{n+1}=\frac{a_{n-1}}n +3 $$

Answer 2

HINT

Assuming you meant $\frac{n+1}{n}$ (since you stated that you factored $n+1$ from the rhs), you can divide by $n+1$ and define $$b_n=\frac{a_n}{n+1}.$$

Now, what is the equation satisfied by $b_n$? Do you know how to solve such an equation?

Answer 3

$$a_{n}=\Big(\frac{1}{n}+n\Big)a_{n-1}+3(n+1)$$ Divide by $$\prod_{k=1}^{n}\Big(\frac{1}{k}+k\Big)$$ To give $$\frac{a_{n}}{\prod_{k=1}^{n}\big(\frac{1}{k}+k\big)}=\frac{a_{n-1}}{\prod_{k=1}^{n-1}\big(\frac{1}{k}+k\big)}+\frac{3(n+1)}{\prod_{k=1}^{n}\big(\frac{1}{k}+k\big)}$$ Let $$A_{n}=\frac{a_{n}}{\prod_{k=1}^{n}\big(\frac{1}{k}+k\big)}$$ To give $$A_{n}-A_{n-1}=f(n)$$ Where $f(n)=\frac{3(n+1)}{\prod_{k=1}^{n}\big(\frac{1}{k}+k\big)}$. Now sum both sides from $2$ to $n$ $$\sum_{k=2}^{n}(A_{k}-A_{k-1})=\sum_{k=2}^{n}f(k)$$ This is $$A_{n}-A_{1}=\sum_{k=2}^{n}f(k)$$ Thus $$A_{n}=A_{1}+\sum_{m=2}^{n}\frac{3(m+1)}{\prod_{k=1}^{m}\big(\frac{1}{k}+k\big)}$$ and $$a_{n}=\Big(\prod_{l=1}^{n}\big(\frac{1}{l}+l\big)\Big)\Big(\frac{a_{1}}{2}+\sum_{m=2}^{n}\frac{3(m+1)}{\prod_{k=1}^{m}\big(\frac{1}{k}+k\big)}\Big)$$