Find an explicit formula for a recursive series by induction.

Question

$$a_1=1,\quad a_{n+1}=\frac{2n+1}{n+1}\cdot a_n$$

find and prove by induction an explicit formula for $a_n,n\in\mathbb{N}$.

The first few elements of the series I discovered were $$1,1.5,2.5,4.375,7.875,...$$

Answer 1

Using telescope, the formula is $$a_n=\frac{(2n+1)!!}{(n+1)!}$$

Answer 2

Given : $$a_{n+1}=\frac{2n+1}{n+1}\cdot a_n$$ and $a_1 = 1$ so, when n = 0 in equation 1 above, $a_1$ will be 1, for$n = 1$ $$a_2 = \frac{3}{2} a_1$$ so, for $n = k$ it will be $$a_{k+1} = \frac{2k + 1}{k +1 }a_{k}$$

now, multiplying both sides of above equation by $\frac{2k + 3}{k+2}$ we will get $$\frac{2k+3}{k+2}a_{k+1} = \frac{2k+3}{k+2}(\frac{2k + 1}{k +1 }a_{k})$$

we know from previous equation that $a_{k+1} = \frac{2k + 1}{k +1 }a_{k}$ therefore we can re-write our equation term $\frac{2(k + 1) + 1}{(k +1 ) + 1}a_{k+1}$ as $a_{(k+1)+1}$ thus, $$a_{(k+1)+1} = \frac{2(k + 1) + 1}{(k +1 ) + 1}a_{(k+1)}$$

Hence, proved thus, the formula would be $a_n = \frac{(2n+1)!}{(n+1)!}$