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I tutored a 10th grader and I was asked this puzzle and I had spent nearly an hour with it and got “no where”. Any one can crack it? Please let me know. Thank you.

Question: Find the $14$ th term of the sequence: $$ \frac{1}{2}, \frac{3}{7}, \frac{1}{3}, \frac{5}{19}, \frac{3}{14}, .... $$.

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  • $\begingroup$ Hmm... I don't think simply tweaking the numbers to generalize the expression will work here... How did you attempt to solve this? I can't find any relations between the terms to call this a sequence, or maybe I don't know something which is required to solve. $\endgroup$
    – rikusp2002
    May 21, 2019 at 3:29
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    $\begingroup$ @SoumalyaPramanik Hint, make the numerators 2,3,4,5,6,... $\endgroup$
    – Mirko
    May 21, 2019 at 4:25

1 Answer 1

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$\displaystyle\frac24\ ,\ \frac37\ ,\ \frac4{12}\ ,\ \frac5{19}\ ,\ \frac6{28}\ ,...$

differences between denominators are $3,5,7,9,...$
(and numerators are consecutive integers $2,3,4,5,6,...$ )

$\displaystyle\frac24,\frac37,\frac4{12},\frac5{19},\frac6{28}, \\ \displaystyle\frac7{39},\frac8{52},\frac9{67},\frac{10}{84},\frac{11}{103},\\ \displaystyle\frac{12}{124},\frac{13}{147},\frac{14}{172},\frac{15}{199},\frac{16}{128}$

elementary watson, $\displaystyle a_{14}=\frac{15}{199}$ .

More generally, $\displaystyle a_n = \frac{n+1}{n^2+3}$ .

Using the formula, $\displaystyle a_{14}=\frac{14+1}{14^2+3}=\frac{14+1}{196+3}=\frac{15}{199}$ .

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    $\begingroup$ Thank you, Mirko. My algebra is rusty and should wear glass to see far... $\endgroup$
    – DeepSea
    May 21, 2019 at 4:26
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    $\begingroup$ Thank you, I am happy I was lucky ... $\endgroup$
    – Mirko
    May 21, 2019 at 4:27
  • $\begingroup$ Excellent, this didn't cross my mind at all 😂 @Mirko $\endgroup$
    – rikusp2002
    May 21, 2019 at 15:39

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