sigma notation

tsh44

Junior Member
Joined
Sep 4, 2005
Messages
67
Hi I have no lcue what the symbol in thsi problem means. I was told to evaluate it.

There's a sigma with a 100 over it and k=1 under it. Then to the right there is a symbol shaped like a "L", then a SQRTk, and then a backwards "L" closing it. It also says "where the L k then backwards L is the greatest integer of k. Pleas ehlp me. I have no idea what it means
 
You did a good job of describing \(\displaystyle \L
\sum\limits_{k = 1}^{100} {\left\lfloor {\sqrt k } \right\rfloor }\).

The notation \(\displaystyle \L
\left\lfloor x \right\rfloor\) is called the ‘floor’ function.
Its value is the greatest integer which does NOT exceed N.

Thus, \(\displaystyle \L
\begin{array}{l}
\left\lfloor 1 \right\rfloor = 1 \\
\left\lfloor {\sqrt 2 } \right\rfloor = 1 \\
\left\lfloor {\sqrt 5 } \right\rfloor = 2 \\
\left\lfloor {\sqrt {10} } \right\rfloor = 3 \\
\end{array}\).

Here is an example: \(\displaystyle \L
\left\lfloor {\sqrt 9 } \right\rfloor + \left\lfloor {\sqrt {10} } \right\rfloor + \left\lfloor {\sqrt {11} } \right\rfloor + \left\lfloor {\sqrt {12} } \right\rfloor + \left\lfloor {\sqrt {13} } \right\rfloor + \left\lfloor {\sqrt {14} } \right\rfloor + \left\lfloor {\sqrt {15} } \right\rfloor = 21\).
Note that there are 7 integers between 8 & 16.
Now you do the sum.
 
Ok I believe I understand the idea of floor function. There are 3 integers that equal 1, 5 that equal 2, 7 that equal three so it goes up by odd numbers. But how would I express this in a rule so i don't have to go through each individual number from 1-100?
 
I got 625 by doing it this way

1-3
2-5
3-7
4-9
5-11
6-13
7-15
8-17
9-19
10-1

I multiplied them and added them all up and got 625 but is there an easier way?
 
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