**Question**

**Introduction**

This is a nasty multiple-choice question to
be set for a primary 6 (~ grade 6) pupil for a test. It could also be a time trap as the pupil
might spend a lot of time to no avail just to try to score that miserable mark. With advanced knowledge, we can solve this using telescoping sums (a.k.a. the method of differences), where many terms cancel and the sum
can be shortened, much like a telescope.

However, this solution looks like over-kill.
Is there a solution that is more
accesible to a primary 6 pupil? Read on!

**A Simpler Solution?**

We observe that a denominator of 5 is
common. In fact, since every term has an
even factor, we have a common factor of 2 in the denominator also. That means every term in the sum can be
expressed as

^{1}/_{10}of something [ Heuristic H09 ]. We can quickly work out the first few partial sums and express each of them as^{1}/_{10}of something. [ Heuristic H04 ]
Tabulating
the results [ H02 & H03 ], we observe
that each partial sum is

^{1}/_{10}of something slightly less than ½. So we try to express them as ½ minus something. How? We can subtract to find out. [ H05 ] For example, if^{5}/_{12}=^{1}/_{2}– ??? then ??? =^{1}/_{2}–^{5}/_{12}=^{1}/_{12}.
Later
on, we notice [
H04 again ] another pattern: the factors in the denominators of the
subtracted quantity matches the last two denominators of the last term. If we considered the full sum, then what
would those factors be? (See the part
highlighted in yellow). These would be
20 and 21, as per the last term of the series.
Using this pattern, we work out the required sum as follows

That is the answer!

H02. Use a diagram / model

H03. Make a systematic list

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H09. Restate the problem in another way

## No comments:

## Post a Comment