Since 64/3 = 21 + 1/3 > 21, I assume <em>S</em> is supposed to be the value of the infinite sum. So we have for some constants <em>a</em> and <em>r</em> (where |<em>r</em> | < 1),
Consider the <em>k</em>-th partial sum of the series,
Multiply both sides by <em>r</em> :
Subtract this from :
Now as <em>k</em> goes to ∞, the <em>r ᵏ</em> term converges to 0, which leaves us with
which we can solve for <em>a</em> :
Meanwhile, the 3rd partial sum is given to be
Substitute <em>a</em> into this equation and solve for <em>r</em> :
Now solve for <em>a</em> :
It follows that