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Vassil Vassilev , a 14 year old Bulgarian student from St Michael's College, Leeds, sent the following solution:
Element No 1 $$ {1\over 2} + {2\over 1} = {1 + 4\over 2} = {5\over 2} = 2 + {1\over 2} = 2 + {1\over 1\times(1+1)} $$ Element No 2 $$ {2\over 3} + {3\over 2} = {4 + 9\over 6} = {13\over 6} = 2 + {1\over 6} = 2 + {1\over 2\times(2+1)} $$ Element No 3 $$ {3\over 4} + {4\over 3} = {9 + 16\over 12} = {25\over 12} = 2 + {1\over 12} = 2 + {1\over 3\times(3 + 1)} $$ Element No 4 $$ {4\over 5} + {5\over 4} = {16 + 15\over 20} = {41\over 20} = 2 + {1\over 20} = 2 + {1\over 4\times(4 + 1)} $$ Element No n
Some, but not all, of the points on the graph of $$ y = x + {1\over x} $$ represent terms of the sequence. You might like to draw the graph and look at what happens to it around $x = 1$.
Vassil worked out the $n$th term of the similar sequence formed by adding the squares of these fractions and after doing some algebra to simplify the expression he obtained the following result: $$ \left({n\over n + 1}\right)^2 + \left({n + 1\over n}\right)^2 = 2 + \left({2n + 1\over n(n + 1)}\right)^2. $$ Here again the limit as n tends to infinity is 2. You might like to prove this result in another way.