Problem 25, section 3
.
2
Ugh. This problem. Well, lets see. We want to find a sum for
m
summationdisplay
k
=0
floorleft
√
k
floorright
For simplicity, we’ll let
n
=
floorleft
√
m
floorright 
1, so that
floorleft
√
m
floorright
=
n
+ 1. Why is
n
significant? Well, if you right out the sum
m
summationdisplay
k
=0
floorleft
√
k
floorright
= 0 + 1 + 1 + 1 + 2 + 2 + 2 + 2 + 2 + 3 + 3 +
. . .
you’ll notice that
n
is the second to last term that appears (before we start
adding
floorleft
√
m
floorright
. This is important because we don’t know how many times we
add
floorleft
√
m
floorright
. But we do now how many times we add everything below it. As you
can see above, this fromula looks a lot like
∑
m
k
=0
(2
k
+ 1)
k
. But is isn’t quite
the same, because we may not add
floorleft
√
m
floorright
as much as possible. Thus, we need
to break this sum into two parts:
n
summationdisplay
i
=0
(2
k
+ 1)
k
+
summationdisplay
floorleft
√
m
floorright
You’ll notice the sum on the left is relatively straight forward, and if you
work it out, it is equal to
n
(
n
+1)(2
n
+1)
6
+
n
(
n
+1)
2
. Thus we only need to figure
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 Fall '08
 Miller
 Math, Addition, Want, Elementary arithmetic, Plus and minus signs, biggest square

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