By knowing, that $a_{n+1}=(n+3)a_n$, how can I find $a_n$? $$a_1=1$$
$$a_{n+1}=(n+3)a_n$$
How can I get to the answer of this, which is:
$$a_n=\frac{(n+2)!}{6}$$
 A: $$a_{n+1}=(n+3)a_n=(n+3)(n+2)a_{n-1}=(n+3)(n+2)(n+1)a_{n-2}=...$$
can you find the pattern now ?
A: write it as follows 
$$\frac { { a }_{ n+1 } }{ { a }_{ n } } =n+3\\ \frac { { a }_{ 2 } }{ { a }_{ 1 } } =4,\frac { { a }_{ 3 } }{ { { a } }_{ 2 } } =5,\frac { { a }_{ 4 } }{ { a }_{ 3 } } =6,...\frac { { a }_{ n } }{ { a }_{ n-1 } } =n+2\\ \\ \frac { { a }_{ 2 } }{ { a }_{ 1 } } \cdot \frac { { a }_{ 3 } }{ { { a } }_{ 2 } } \cdot \frac { { a }_{ 4 } }{ { a }_{ 3 } } \cdot ...\cdot \frac { { a }_{ n } }{ { a }_{ n-1 } } =\frac { \left( n+2 \right) ! }{ 6 } $$
A: Upon taking logarithms we have,
$$\ln a_{n+1}=\ln a_n+\ln (n+3)$$
Little Algebra and introducing dummy variable.
$$\ln  a_{i+1}-\ln a_{i}=\ln (i+3)$$
Summing both sides from $i=1$ to $n-1$ and noticing that the sum is a telescoping sum,
$$\ln  a_{n}-\ln 1=\sum_{i=1}^{n-1} \ln (i+3)$$
Logarithmic rules,
$$\ln a_{n}=\ln \prod_{i=1}^{n-1} (i+3)$$
$$a_n=\prod_{i=1}^{n-1} (i+3)$$
Note the $(1)(2)(3)$ we would like is missing.
$$3! a_n=(1)(2)(3)\prod_{i=1}^{n-1} (i+3)$$
Noting definition of factorial.
$$3! a_n=(n+2)!$$
Solving and simplifying,
$$a_n=\frac{(n+2)!}{6}$$
A: Let ${\{a_n\}}_{n = 1}^{\infty}$ be the number sequence
$$
a_1 = 1 \qquad \mbox{ and } \qquad a_{n + 1} = (n + 3) a_n \quad \mbox{ for all } \quad n = 1 , 2 , \ldots
$$
and let ${\{b_n\}}_{n = 1}^{\infty}$ be the number sequence
$$
b_n = \frac{(n + 2) !}{6} \quad \mbox{ for all } \quad n = 1 , 2 , \ldots\mbox{,}
$$
so let's show that $a_n = b_n$ for all $n = 1 , 2 , \ldots$ and we do using induction method. On the one hand,
$$
a_1 = 1 = \frac{6}{6} = \frac{(1 + 2) !}{6} = b_1\mbox{.}
$$
On the other hand, let $n$ be a natural number, with $n > 1$, let's suppose that $a_n = b_n$ and let's show the equality $a_{n + 1} = b_{n + 1}$. It isn't difficult:
$$
a_{n + 1} = (n + 3) a_n = (n + 3) b_n = (n + 3) \frac{(n + 2) !}{6} = \frac{(n + 3) !}{6} = \frac{((n + 1) + 2) !}{6} = b_{n + 1}\mbox{.}
$$
