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${1}^{2}-{2}^{2}+{3}^{2}-{4}^{2}+\cdots +{1999}^{2}$19992 using closed form formula for sum of squares of first n natural numbers
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opiftinkistj6mi6
If you first add all the squares:
${1}^{2}+{2}^{2}+{3}^{2}+{4}^{2}+{5}^{2}+{6}^{2}+\cdots$
and then subtract the even squares:
$\phantom{{1}^{2}}-{2}^{2}\phantom{+{3}^{2}}-{4}^{2}\phantom{+{5}^{2}}-{6}^{2}\phantom{-}\cdots$
then you are left with only the odd squares:
${1}^{2}\phantom{+{2}^{2}}+{3}^{2}\phantom{+{4}^{2}}+{5}^{2}\phantom{+{6}^{2}}+\cdots$
You need to subtract the even squares one more time to get
${1}^{2}-{2}^{2}+{3}^{2}-{4}^{2}+{5}^{2}-{6}^{2}+\cdots$
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fulgoripb0b
We can use the identity
$\left(x+1{\right)}^{2}-{x}^{2}=x+1+x$
to tackle the sum in 2 simpler ways:
Way 1
$\begin{array}{rl}& {1}^{2}-{2}^{2}+{3}^{2}-{4}^{2}+{5}^{2}-...-{1998}^{2}+{1999}^{2}\\ =& 1+\left(2+3\right)+\left(4+5\right)+\dots +\left(1998+1999\right)\\ =& 19990000\end{array}$
Way 2
$\begin{array}{rl}& \left({1}^{2}-{2}^{2}+{3}^{2}-{4}^{2}+{5}^{2}-{6}^{2}...-{1998}^{2}\right)+{1999}^{2}\\ =& -\left(\left(1+2\right)+\left(3+4\right)+\left(5+6\right)+\dots +\left(1997+1998\right)\right)+{1999}^{2}\\ =& 19990000\end{array}$