Number pattern

$\begin{array}{cl} &1^2-2^2+3^2-4^2+5^2-\cdots-2010^2+2011^2\\ =&1+(3^2-2^2)+(5^2-4^2)+\cdots+(2011^2-2010^2)\\ =&1+(3-2)(3+2)+(5-4)(5+4)+\cdots+(2011-2010)(2011+2010)\\ =&1+3+2+5+4+\cdots+2011+2010\\ =&1+2+3+4+5+\cdots+2010+2011\\ =&\dfrac{2011\times2012}{2}\\ =&2023066 \end{array}$

There are 1006 odd, and 1005 even terms. The formulas for generating odd and even squares are: $\frac{n(2n-1)(2n+1)}{3}$ and $\frac{2n(n+1)(2n+1)}{3}$ respectively. Hence you can subtract the two from each other and then simplify to get: (1006)(2011)(3)/3 = (1006)(2011) = 2023066

$(1-2)(1+2)+(3-4)(3+4)+ ... +(2009-2010)(2009+2010)+2011^{2}$ $=-(1+2+3+4+ ... + 2010) +2011^{2}$ $=2011(2011-1005)=2023066$

The series can be written as

(1^2-2^2) + (3^2-4^2) + (5^2-6^2)......... +(2009^2-2010^2) +2011^2

Now use a^2-b^2 = (a-b)(a+b)

=(-1)(3) + (-1)(7) + (-1)(11).......+(-1)(4019) +2011^2

= -3 -7 - 11 -............-4019 +2011^2

= -(3+7+11.....4019) + 2011^2

Now, the series in the brackets is in AP, hence sum = n/2(a+l)

= - (1005/2(4022) +2011^2

= -(1005)(2011) + 2011(2011)

= 2011(2011-1005) = 2011 X 1066 = 2023066

$\displaystyle \sum^{1005}_{n=0}(2n+1)^2-\displaystyle \sum^{1005}_{n=1}(2n)^2 \\ 1^2+ \displaystyle \sum^{1005}_{n=1}(4n^2+4n+1)- \displaystyle \sum^{1005}_{n=1}(4n^2) \\ 1+\cancel{4 \displaystyle \sum^{1005}_{n=1}n^2}+4 \displaystyle \sum^{1005}_{n=1}n+ \displaystyle \sum^{1005}_{n=1}1-\cancel{4 \displaystyle \sum^{1005}_{n=1}n^2} \\ 1+4\cdot \frac{1005×1006}{2}+1005=\boxed{2023066}$

$1^2 - 2^2 + 3^2 - 4^2 + ... - 2010^2 +2011^2 =$ $= 1^2 + 2^2 + 3^2 + ... + 2011^2 - 2(2^2 + 4^2 + 6^2 + ... + 2011^2) =$ $= 1^2 + 2^2 + 3^2 + ... + 2011^2 - 8(1^2 + 2^2 + 3^2 + ... + 1005^2) =$ $= \frac {2011 \cdot 2012 \cdot 4023}{6} - 8(\frac {1005 \cdot 1006 \cdot 2011}{6}) = \boxed{2023066}$