## Challenge #4 Solution, Part II

But wait! What happened to Part I, you ask!? Well, let’s play the Multiple Choice Game!

(a) There is no Part I.
(b) There USED to be a Part I, but it disappeared.
(c) Part I will come later. Haven’t you seen the Star Wars movies?
(d) Quit whining.
(e) All of the above.

And the answer is… actually, before I tell you the answer, let me just mention that the only reason anyone ever gives multiple choice tests is because they’re easy to grade, not because they make for a good test. But I digress.

Remember the second question from Challenge #4, way back when? Of course you do, but here it is again so you don’t have to tire your finger clicking on that link. The challenge was to find the sum of the first 100 terms of this series:

$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \cdots$.

Hmm… first, of course, we have to figure out the pattern. Do you see it? Just looking at the denominators, you might notice that the differences between successive denominators go like this: 4, 6, 8, 10, 12… in other words, successive even numbers. So the next denominator would be 42 + 14 = 56, then 56 + 16 = 72, and so on. OR you might notice that 2 = 1×2, 6 = 2×3, 12 = 3×4, 20 = 4×5, and so on… so the next denominator would be 7×8 = 56, then 8×9 = 72… huh! Isn’t that strange that two different (and seemingly unrelated) rules give the same sequence of numbers? Well… it isn’t strange, actually, but I’ll leave it up to anyone interested to show why these two seemingly unrelated rules yield the same number sequence. (If you know about arithmetic series it shouldn’t be too hard!)

Now that we know the pattern for the terms of the series, let’s add up the first two terms, then add up the first three terms, then four, five, and so on, to see if we notice a pattern.

$\begin{array}{rcl} \frac{1}{2} + \frac{1}{6} & = & \frac{2}{3} \\ \frac{1}{2} + \frac{1}{6} + \frac{1}{12} & = & \frac{3}{4} \\ \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} & = & \frac{4}{5} \\ \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} & = & \frac{5}{6} \end{array}$

I think you probably see the pattern, don’t you? Will this always be true? It’s a good guess… and it turns out it’s not too hard to prove it by induction. What’s that, you ask? Never fear, that’s a subject for another post soon. Assuming for now that this pattern continues to hold true, the sum of the first 100 terms is of course… 100/101.