How do I find the general term for the following recursive sequence

Question

$a_{n+1} = -\frac{5a_n}{2n+1}$ for $n = 1,2,3,\dots$

Ex. For $a_1 = 1$

$a_2 = -\frac{5}{3}$

$a_3 = \frac{5}{3}$

$a_4 = -\frac{25}{21}$

$a_5 = \frac{125}{189}$

$a_6 = -\frac{625}{2079}$

etc.

Thank you

Answer 1

Let's try to expand explicitly a few terms : $$ a_n = \frac{-5}{2n-1} a_{n-1} = \frac{-5}{2n-1} \frac{-5}{2n-3} a_{n-2} = \frac{-5}{2n-1} \frac{-5}{2n-3} \frac{-5}{2n-5} a_{n-3} $$ So it is rather clear that the numerator will be of the form $5^k$ and the denominator of the form "$1 \cdot 3 \cdot 5 \cdot 7$". Also note that because of the minus sign, the sign of the sequence will flip every time. So we will also have a factor of the form $(-1)^k$. Now let's observe a bit the first few terms : $$ a_1 = 1 \\ a_2 = \frac{-5}{3} \\ a_3 = \frac{25}{3 \cdot 5} \\ a_4 = \frac{-125}{3 \cdot 5 \cdot 7} $$ We can conjecture that $$a_n = (-1)^{n+1} \frac{5^{n-1}}{\prod_{i=1}^{n-1} (2i+1)}$$ (Where the empty product is defined to be $1$ when $n=1$)