Sequence

Let's do this a different way! 
Group the terms in pairs. 
1000² - 999² = (1000+999)(1000-999) = 1999. 
998² - 997² = (998 + 997)(998-997) = 1995. 
. 
. 
. 
2²-1² = (2+1)(2-1) = 3. 

So we have to sum the arithmetic progression 
3 + 7 + ... + (4n-1) 
to 500 terms, since 1999 is the 500th term of 
this progression. 
So, how do we do this? 
Note that 
3 + 1999 = 2002 
7 + 1995 = 2002 
11 + 1991 = 2002 
. 
. 
. 
1999 + 3 = 2002 
Call the sum of each column S 
Then S + S = 500*2002 
S = 500*1001 = 500500.

1000^2 + 999^2 - 998^2 - 997^2..........etc.

Why did you subtract the first 2 terms?

Every 4 terms form an arithmetic sequence as follows:
7,988, 7,956, 7,924, 7,892........etc., where:
First term =7,988
Common difference = - 32
Number of terms =1,000 / 4 = 250
SUM = N/2 * [2*F + (N - 1)* D]
         =250/2*[2*7,988 + (250 - 1) * - 32]
         =125 * [15,976 + (249 * - 32)]
         =125 * [15,976 - 7,968]
         =125 * [8,008]
         = 1,001,000

∑[- 4(8n - 2005), n, 1, 250] = 1,001,000

What is the sum of this sequence?
1000^2 + 999^2 - 998^2 - 997^2 +996^2 + 995^2 - .........+ 4^2 + 3^2 - 2^2 - 1^2.

\(\begin{array}{|rcll|} \hline && \mathbf{1000^2 + 999^2 - 998^2 - 997^2 +996^2 + 995^2 - \ldots + 4^2 + 3^2 - 2^2 - 1^2} \\ &=& \sum \limits_{k = 1}^{250} \Big( (4k)^2+ (4k-1)^2- (4k-2)^2 -(4k-3)^2 \Big) \\ &=& \sum \limits_{k = 1}^{250} \Big( 16k^2 +16k^2-8k+1-16k^2+16k-4 -16k^2+24k-9 \Big) \\ &=& \sum \limits_{k = 1}^{250} \Big( -8k+1 +16k-4 +24k-9 \Big) \\ &=& \sum \limits_{k = 1}^{250} \Big( 32k-12 \Big) \\ &=& \sum \limits_{k = 1}^{250}32k-\sum \limits_{k = 1}^{250}12 \\ &=& 32\sum \limits_{k = 1}^{250}k- 12\cdot 250 \\ &=& 32\left(\dfrac{1+250}{2}\right)\cdot 250- 12\cdot 250 \\ &=& 16\cdot 251 \cdot 250- 12\cdot 250 \\ &=& 250( 16\cdot 251 - 12) \\ &=& 250\cdot 4004 \\ &\mathbf{=}& \mathbf{1001000} \\ \hline \end{array}\)