A Tricky Equation

Solution

Let S(n) = 13 + 2³ + ... + n³

Then:

2³ + 4³ + ... + (2n)³
= 2³(1³ + 2³ + ... + n³)
= 2³S(n)

Also:

1³ + 3³ + ... + (2n-1)³
= 1³ + 2³ + ... + (2n)³ - (2³ + 4³ + ... + (2n)³)
= S(2n) - 2³S(n)

Now we use the fact that S(n) = [n(n+1)/2]², which can easily be proved by induction.

S(2n) = [(2n)(2n+1)/2]² = n²(2n+1)²
2³S(n) = 2³[n(n+1)/2]² = 2n²(n+1)²

And so the left hand side of the original equation reduces to:

(n²(2n+1)² - 2n²(n+1)²) / 2n²(n+1)²
= ((2n+1)² - 2(n+1)²) / 2(n+1)²
= (2n² - 1) / (2n² + 4n + 2)

Setting this equals to 199/242 gives

484n² - 242 = 398n² + 796n + 398
86n² - 796n - 640 = 0
(86n + 64)(n - 10) = 0

So n = 10.