Sum of series

1^2 - 2^2 + 3^2 - 4^2 .......... - 100^2

1 - 4 + 9 - 16 + 25 -36.. ... ... .... .... - 100^2

-3 + -7 + -11..... ............

so we get aritmathic problem n = 50 because we allready use 2 to make arithmathic

so, SUM = n/2 (2a + (n-1)b) a = first term b= gap beetwen first term and second term

SUm = 50/2 (2(-3) + (50-1) -4) = 25 ( -6 + -196) = 25 ( -202) = -5050

the series is 1, -3, 6, -10, 15, -21, 28, 36, -45 The sequence is of triangular numbers given by n(n+1)/2 where n is the nth term of series n=100. 100*101/2= 5050 finally every even term will yield negative value so -5050

$1^2 - 2^2 + 3^2 - 4^2 + ... + 99^2 - 100^2 =$

$(1 - 2)(1 + 2) + (3 - 4)(3 + 4) + ... + (99 - 100)(99 + 100) =$ Difference of squares

$(-1)(1 + 2) + (-1)(3 + 4) + ... + (-1)(99 + 100) =$

$-(1 + 2 + 3 + 4 + ... + 99 + 100) =$ Factor out -1

$- \frac{100}{2}(1 + 100) =$ Arithmetic sum formula

$- 50(101) =$

$\boxed{-5050}$

sum(1^2+2^2+3^2+...+n^2)= n (n+1) (2n+1)/6 (>>>this is a well known fact) So, here, the series is: 1^2-2^2+3^2+...+99^2-100^2 which can be broken into: (1^2+2^2+3^2+...+100^2) - 2 * (2^2+4^2+6^2+...+100^2) =(1^2+2^2+3^2+...+100^2) - 2 * 4 * (1^2+2^2+3^2+...+50^2) (>>>taking out 2^2 common from second bracket) =(100 * 101 * 201)/6 - (8 * 50 * 51 * 101)/6 =50 * 101 * (67-68) (>>>simplification, after geting common factors from both fractions) = (-5050), that is the answer :)

DISCLAIMER : This is a whatever-gets-the-job-done approach. As I am not really into technical (for the lack of a good term) mathematics, what I offer here is a long roundabout method. There definitely are shorter solutions than this one. I am only sharing how I found the answer.

The numbers' squares proceed as

$1, 4, 9, 16, 25, \ldots$

The differences of all $(i + 1)^2 - i^2$ form the sequence of odd numbers:

$1 = 1 = n^1$

$3 = 4 - 1 = n^2 - n^1$

$5 = 9 - 4 = n^3 - n^2$

$7 = 16 - 9 = n^4 - n^3$

$9 = 25 - 16 = n^5 - n^4 \ldots$

and the sequence can be thought of as

$1, (1^2 + 3), (2^2 + 5), (3^2 + 7), (4^2 + 9), \ldots$

Take $5^2$ . $5^2$ is the sum of $4^2$ and $9$ . But $4^2 = 3^2 + 7$ , $3^2 = 2^2 + 5$ , and $2^2 = 1^2 + 3$ . The same thing can be said for all $n^2$ . Thus, the recent sequence above can be rewritten as

$1, (1 + 3), (1 + 3 + 5), (1 + 3 + 5 + 7), (1 + 3 + 5 + 7 + 9), \ldots$

Now, we consider the given problem but I will now state it as:

$1 - (1 + 3) + (1 + 3 + 5) - (1 + 3 + 5 + 7) + (1 + 3 + 5 + 7 + 9) - \ldots$

For instance, we have the sequence until $6^2$ . If the numbers are regrouped and added, we get

$1 - (1 + 3) + (1 + 3 + 5) - (1 + 3 + 5 + 7) + (1 + 3 + 5 + 7 + 9) - (1 + 3 + 5 + 7 + 9 + 11)$

$= 1 - (1) - 3 + (1 + 3 + 5) - (1 + 3 + 5) - 7 + (1 + 3 + 5 + 7 + 9) - (1 + 3 + 5 + 7 + 9) + 11$

$= 3 + 7 + 11$

If we only considered the sum up to $2^2$ , we only have $3$ . If up to $4^2$ , we have $3 + 7$ . For up to $6^2$ , as we saw just above, the sum is $3 + 7 + 11$ . We have the sum of a sequence of numbers with $a_1 = 3$ and $a_{i + 1} - a_i = d = 4$ . With the original question having terms up to $100^2$ , we can regroup the sequence as fifty pairs of $(n + 1)^2 - n^2$ . The sum of this sequence is

$S_n = \frac{n}{2}(2a_1 + (n - 1)(d)) = \frac{50}{2}(2(3) + (50 - 1)(4)) = \boxed{5050}.$