1. sum problem

I posted before several weeks this problem
Determine the following sum:
1-4+9-16+25-...-10000+10201

i got to solutions from the forum members
1 - $\displaystyle S \;=\;\sum^{101}_{k=1}k^2 - 2\sum^{50}_{k=1}(2k)^2 \;=\;\sum^{101}_{k=1}k^2 - 8\sum^{50}_{k=1}k^2$
2- $\displaystyle \sum_{n=1}^{51}\{(2n-1)^2\} - \sum_{i=1}^{50} \{(2i)^2\}$

i want to know how to get the sigma notation
is it by guessing or what?

2. Originally Posted by Abbas
I posted before several weeks this problem
Determine the following sum:
1-4+9-16+25-...-10000+10201

i got to solutions from the forum members
1 - $\displaystyle S \;=\;\sum^{101}_{k=1}k^2 - 2\sum^{50}_{k=1}(2k)^2 \;=\;\sum^{101}_{k=1}k^2 - 8\sum^{50}_{k=1}k^2$
2- $\displaystyle \sum_{n=1}^{51}\{(2n-1)^2\} - \sum_{i=1}^{50} \{(2i)^2\}$

i want to know how to get the sigma notation
is it by guessing or what?
lets take this:

1-4+9-16+25-...-10000+10201

and seperate it into:

(1+9+25+....+10201) - (4+16+......+10000)

which is just:

(sum of squares of odd numbers) - (sum of squares of even numbers)

which leads to the second solution...& we should sum it up to 101 & 100 respectively as the last terms of the first & second parts are (101)^2 & (100)^2 sp (2n-1)=101 & (2n)=100

(and thats exactly how you get the sum as you get n=51 for the first & n=50 for the second)

$\displaystyle \sum_{n=1}^{51}\{(2n-1)^2\} - \sum_{n=1}^{50} \{(2n)^2\}$

3. is it right mathematically to seprate them like this?
u'll expand the first part ( complete square ) and the find the sum and subtract with the other sum?
thanx alot

4. it is certainly. as it will lead us back to:

(1+9+25+....+10201) - (4+16+......+10000)
(as we are subtracting the sums)

which again is:

(1-4+6-16+.................)

which is your problem. hope you got it.