2860.41 – Fraction Pattern


Extend this pattern and generalize it:

1+35+7=13\frac{1+3}{5+7} = \frac{1}{3}

1+3+57+9+11=13\frac{1+3+5}{7+9+11} = \frac{1}{3}

1+3+5+79+11+13+15=13\frac{1+3+5+7}{9+11+13+15} = \frac{1}{3}


Solution

We use the fact that the sum of the first nn odd numbers is n2n^2.

E.g. 1+3+5+7=16=421 + 3 + 5 + 7 = 16 = 4^2.

We also use the fact (this is tricky) that, say, 7+9+117 + 9 + 11 can be thought of as 1+3+5+7+9+11(1+3+5)1 + 3 + 5 + 7 + 9 + 11 - (1 + 3 + 5), i.e. (6232)(6^2-3^2).

Thus the denominator in each of these fractions is the sum of the first 2n2n odd numbers minus the sum of the first nn odd numbers:

1+35+7=1+31+3+5+7(1+3)=224222=4164=412=13\frac{1+3}{5+7} = \frac{1+3}{1+3+5+7-(1+3)} = \frac{2^2}{4^2-2^2} = \frac{4}{16-4} = \frac{4}{12} = \frac{1}{3}. Aha!

And 1+3+5+79+11+13+15=1+3+5+71+3+5+7+9+11+13+15(1+3+5+7)=428242=166416=1648=13\frac{1+3+5+7}{9+11+13+15} = \frac{1+3+5+7}{1+3+5+7+9+11+13+15-(1+3+5+7)} = \frac{4^2}{8^2-4^2} = \frac{16}{64-16} = \frac{16}{48} = \frac{1}{3} YES!

We are heading toward the formula n2(2n)2n2=n24n2n2=n23n2=13\frac{n^2}{(2n)^2-n^2} = \frac{n^2}{4n^2-n^2} = \frac{n^2}{3n^2} = \frac{1}{3}.