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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: $$ \frac{1}{2}, \frac{3}{7}, \frac{1}{3}, \frac{5}{19}, \frac{3}{14}, .... $$.

rikusp2002
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DeepSea
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  • 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. – rikusp2002 May 21 '19 at 03:29
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    @SoumalyaPramanik Hint, make the numerators 2,3,4,5,6,... – Mirko May 21 '19 at 04:25

1 Answers1

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

Mirko
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