Is the Series S = 12-22+32-42...+20092 Equivalent to -(1+2+3+...+2008)?

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The discussion revolves around the equation S = 12 - 22 + 32 - 42 + ... + 20092 and whether it can be expressed as -(1 + 2 + 3 + ... + 2008). Participants confirm that this transformation is valid, explaining that each pair of terms can be factored to yield negative sums of natural numbers. The conversation also touches on the need to include the term 20092 after factoring. Additionally, hints are provided for calculating the sum of the first n natural numbers, emphasizing the pairing method for simplification. The thread concludes with a focus on understanding summation techniques.
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S = 12-22+32-42...+20092

Attempt=

S = (1+2)(1-2)+(3+4)(3-4)+...+(2007-2008)(2007+2008) [can we write this as -(1+2+3+4+5...2008) if yes, then why ?) +20092
Stuck after this.
 
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Do you know the summation of:
12+22+32+42+...+n2

That would be of help here.
 
Kartik. said:
S = 12-22+32-42...+20092

Attempt=

S = (1+2)(1-2)+(3+4)(3-4)+...+(2007-2008)(2007+2008) [can we write this as -(1+2+3+4+5...2008) if yes, then why ?) +20092
Stuck after this.

Yes you can write it as -(1+2+...+2008)+20092 because

(1+2)(1-2) = -(1+2)
(3+4)(3-4) = -(3+4)
...

And finally, we factored 20072-20082 but left out the 20092 term so we need to add that term in at the end.

Now, do you know the formula to sum the first n natural numbers?

Hint:

1+2+3+...+(n-2)+(n-1)+n

= (1+n) + (2+(n-1)) + (3+(n-2))+...
 
Just to give you a greater hint to support Mentallic, draw out the numbers from 1 to n
1+2+3+4...n-3+n-2+n-1+n, now notice, like Mentallic said, how 1+n = n-1+2 = n-2+3, so all those pairs have the same value, and when you have x things with value y, the result is...
Bonaparte
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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