So my friend gave me this sequence
$\frac 32$ $\quad$ $ \frac 54$ $\quad$ $\frac {21}{16}$ $\quad$ $\frac {45}{32}$
Each of these numbers corresponds to n = 2 , n = 4 , n = 6 ... so to even n values , for odd values of n he gave me nothing , it is blank.
For the denominators i was planning on using $\frac {2^n}{n}$ but the term 8 is missing from this geometric sequence.
For the numerators $ n(n-2)- 3 $ was the plan , however it does not work the first one
I think he might be messing with me , but is there any possible way to find the formula for the nth term of this sequence???
Thank you very much in advance your time and help!
Consider the products \begin{eqnarray*} \frac{1}{1} ,\frac{3}{2} , \frac{3 \times 5}{3 \times 4} ,\frac{3 \times 5 \times 7}{4 \times 5 \times 6} ,\frac{3 \times 5 \times 7 \times 9}{ 5 \times 6 \times 7 \times 8} , \frac{3 \times 5 \times 7 \times 9 \times 11}{ 6 \times 7 \times 8 \times 9 \times 10}, \cdots \end{eqnarray*} Now multiply these by the second to last term \begin{eqnarray*} \frac{1}{1} ,\frac{3}{2} , \frac{ 5}{ 4} ,\frac{7}{8} \times \frac{3}{2} ,\frac{ 9}{ 8} \times \frac{ 5}{ 4}, \frac{ 11}{ 16} \times \frac{21}{16}, \cdots \end{eqnarray*}