$\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 $

  • 4
    $\begingroup$ I like that you put good effort into the question, however as a small warning: this community usually does not like these types of questions, since there in theory could be an infinite amount of answers, and there is no way to determine truly what the "correct" is. $\endgroup$ – Eff Mar 4 '15 at 17:48
  • $\begingroup$ @eff: But i already mentioned the specific answer in the description. $\endgroup$ – R K Mar 4 '15 at 17:49
  • 1
    $\begingroup$ @RK. Let $f:\mathbb{N}\to\mathbb{R}$ be the function defined as $f(1)=\frac73,f(2)=\frac{35}6,f(3)=\frac{121}{12},f(4)=\frac{335}{36}$ and $f(x)=0$ for all $x\ge5$, so answer is $0$. How you can say that it is not true? $\endgroup$ – user164524 Mar 4 '15 at 17:52
  • $\begingroup$ Yes, but there could be any number of justifications for it, and there could be any number of justifications for any other answer, that's my point. What should we do, if we didn't know the "correct" answer? :-) $\endgroup$ – Eff Mar 4 '15 at 17:53
  • $\begingroup$ @Mathematician171 from mathematics point of view i agree , but i think there are some 'good functions' that might exist, mse should be a bit flexible on this , if they dont wanna try fair enough , but downvotes should not be encouraged to these questions $\endgroup$ – avz2611 Mar 4 '15 at 17:55

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 $(12-1)^2=121$.

  • $\begingroup$ Thanks,Ur solution does make sense , not sure if the question was wrong, but as it strongly concludes the expected answer of $\frac{865}{48}$ , i assume the question had the typo. $\endgroup$ – R K Mar 5 '15 at 20:57

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.


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..


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