MATHEMATICAL RECREATIONS 209

491. The distribution of the coconuts will be as follows:
The monkeys get # coconuts, one from each man.
The first man gets y: + 2 coconuts

“ second yo + 2
“  third Y3 + 2
“ nth man Y + 2

The sum of these numbers is

n+ @ity tys+ oo+ ) -0z

If we add the » + 1 equations that we started with in § 488
and cancel and transpose properly, we find that the above
sum equals x, the whole number of coconuts, as it should,
that is,

x=n+O+y+ys+ -+ )+ nz

492. We may use this formula to further check the three
formulas

X = q@mtt —n® =g 54
v = m»*(m + 1)¥(gn + 1) — 1
g=m—D[m+ 1) (gn+1) —1]:n

of § 490.
From the second of these formulas

(1 + )/ +32) = (m + 1)/m

so that the valuesof 1 + y;, 1 + v, 1 + y3, -++, 1 4+ v, form
a geometric progression whose common ratio is (m + 1)/m, as
they should by § 490.

The first term of this progression, 1 + y1, equals m"*(gn + 1)
by the formula. The number of terms is 7.

Hence their sum is [(m + 1)* — m"](gn + 1).

That is,

A4+ +Q+y)+Q+y)+ -+ 1+ y.)
=[m+1)"—m"](gn+ 1)