Lots of Squares

Group with parentheses:

$(1^2 - 3^2) + (5^2 - 7^2) ... + (97^2 - 99^2)$

Evaluate each group of parentheses:

$(-8) + (-24) + (-40) ... + (-392)$

Factor out $-8$ . We know each term must be a multiple of $-8$ because $x^2 - (x + 2)^2 = -4(x + 1)$ , and since $x$ is always odd, $x + 1$ is always even, so $-4(x + 1)$ must always be a multiple of $-8$ .

$-8(1 + 3 + 5 ... + 49)$

The sum of the odd numbers from 1 to 49 is $\frac {50*25}{2}$ or $625$ .

$-8 * 625 = \boxed{-5000}$

Obviously it's sum and difference of squares. Try to factor out.. $(1^{2}- 3^{2}) + (5^{2} - 7^{2}) + (9^{2} -11^{2}) ... + (97^{2} - 99^{2})$

$(1-3)(1+3) + (5-7)(5+7) + (9-11)(9+11) ... + (97-99)(97+99)$

$(-2) \times [ 4 + 12 + 20... + 196 ]$

$(-8) \times [ 1 + 3 + 5... +49]$

For the odd series..

$a_{n} = 2n + 1$

$49 = 2n + 1$

$n = 25$ ---> the nth term of the series

For the sum of odd series

$S_{n} = [\frac{n}{2}](2a_{1} + (n-1)d)$

$S_{n} = [\frac{25}{2}]((2)(1) + (25 - 1)(2))$

$S_{n} = 625$

therefore:

$-8 \times 625 = \boxed{-5000}$

i made pairs of 99-1 , 97-3..............so on.........and calculated using A.P summation

1^2-3^2+5^2-7^2+...+97^2-99^2

= (1-3)(1+3)+(5-7)(5+7)+...+(97-99)(97+99)

= -2(4+12+...+196)

= -2(4+196)[(196-4)/8+1]

= -5000

The answer is simply {The negative difference of the [(negative of 49th summation of (unbounded) 9409 (where n is equivalent to 1) and 9784] = (-5000)} where it can be clearly stated as -5000 which is the correct answer !

1^2-3^2 + 5^2-7^2 ... + 97^2-99^2 (1+3)(1-3) + (5+7)(5-7) ... + (97+99)(97-99) (4)(-2) + (12)(-2) ... + (196)(-2) -2[(4) + (12) ... + (196)] The sum of 4,12 ..., 196 (25 terms) S = n/2 (1st term + Last term) where n = number of terms S = 25/2 (4+196) S = 25/2 (200) S = 25 (100) S = 2500

-2(2500) -5000

1^2 - 3^2 = -2(4)

5^2 - 7^2 = -2(12)

9^2 - 11^2 = -2(20)

...............................

97^2 - 99^2 = -2(196)

Adding

The given number = -2(4 + 12 + 20 + ........... + 196) = -2(sum of arithmetic sequence)

Then

The given number = -2(2500) = - 5000

Group in parentheses: $(1^{2} - 3^{2}) + (5^{2} - 7^{2}) + ... (97^{2} - 99^{2})$

Each parenthesis is a difference of two squares, so you can rewrite it as: $(1-3)(1+3) + (5-7)(5+7) + ... (97-99)(97+99)$

Notice that for each product, the factor that involves a subtraction always equals -2 (1-3=-2, 5-7=-2, 97-99=-2), so you can rewrite it as: $-2(1+3) + -2(5+7) + ... -2(97+99)$

You can factor out a -2: $-2 (1 + 3 + 5 + 7 + ... 97 + 99)$

The term in parenthesis is a simple arithmetic series where $a_{1} = 1, n=50, a_{50}=99$ So that whole series equals $\frac{50(1+99)}{2} = \frac{50(100)}{2} = \frac{5000}{2} = 2500$ Replace (1+3+5+7+... 97+99) with 2500 and:

$-2(2500) = \boxed{-5000}$