1956 AHSME Problems/Problem 25
Problem 25
The sum of all numbers of the form , where
takes on integral values from
to
is:
Solution
The sum of the odd integers from
to
is
. However, in this problem, the sum is instead
, starting at
rather than
. To rewrite this, we note that
is
less than
for every
we add, so for
's, we subtract
, giving us
,which factors as
.
See Also
1956 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.