Find the next term in the sequence. $\frac{7}{3},\frac{35}{6},\frac{121}{12},\frac{335}{36},\ldots $

Question

$\dfrac{7}{3},\dfrac{35}{6},\dfrac{121}{12},\dfrac{335}{36},\ldots $

$\bf\text{Answer}$ given is $\dfrac{865}{48}$

I found that $4^{th}$ differencess of the numbers $7,35,121,335\cdots$ are not constant .

and the second differences of the denominator drastically changes,

$3\quad 6\quad 12\quad 36\quad 48\\~\\ \quad 3\quad 6\quad 24\quad \color{red}{12}$

decimal value is also not showing any pattern.

$\frac{7}{3},\ \frac{35}{6},\ \frac{121}{12},\ \frac{335}{36},\ldots $

$2.33,\ 5.83,\ 10.08,\ 9.33,\ldots $

Answer 1

If your fourth term is $\dfrac{335}{\color{red}{24}}$ instead of $~\dfrac{335}{\color{red}{36}}~,~$ then the pattern you're looking for is

a recurrence relation of the form $~a_{n+1}~=~6n+1-\dfrac{a_n}2~,$ with $~a_1=\dfrac73~.$ This idea

came to me while trying to approximate each term with its nearest integer; in particular,

by noticing that $~6^2=36\simeq35$, and $(12-1)^2=121$.

Answer 2

For the denominators, if we consider skipping terms, 12/3 = 4, so the one past 36 could be 12*4 = 48.

For the numerators, I don't know.

Answer 3

Okay so after solving a lot this is how the answer goes.. The series is 2+1/3 , 6-1/6 , 10+1/12 , 14-1/24 , 18+1/48.. Both the integral part and fractional part follow a specific sequence. Hope This Solved Your Problem..