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# What is the sum of 1000^2-999^2+998^2-997^2+...+2^2-1^2?

The signs alternate between squares of consecutive integers. What's the sum

### 10 Answers

- ironduke8159Lv 71 decade agoFavorite Answer
1000^2-999^2 =(1000-999)(1000 +999) = 1(1999) =1999

998^2-997^2 (998-997)(998+997)= 1995

996^2-995^2=(996-995)(996+995 = 1991

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

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

4^2-3^2= (4-3)(4+3)=7

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

so this is a simple arithmetic sequence starting with 3 the first term and each succesive term is 4 mor than the previous term. There are 500 such terms.

The sum is given by s = .5n[2a1 +d(n-1)]

sum = . 5*500[2*3+ 4(500-1)]

=250[ 6+1996]= 250* 2002=500,500

- steiner1745Lv 71 decade ago
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.

- Anonymous1 decade ago
I assume this goes all the way down to 1^2 right?

You can simply group the expression into binomials:

(1000^2 - 999^2) + (998^2 - 997^2) + ....

Now just find the values of those grouped terms and you'll notice a linear pattern.

Now create an equation to find the last term:

x,y -> 1,1999

2003-4x=y

Now since there are 500 groups there are 500 terms:

2003 - 4(500) = 3

Now all you need to do is find the sum of this which will be the average of the first term (1999) and the last term (3) multiplied by the amount of terms you have (500). This should give you:

500500

- teekshi33Lv 41 decade ago
Look for a pattern..

1^2=1

2^2-1^2=3

3^2-2^2+1^2=6

These are called triangular numbers. The formula is [n(n+1)]/2

with n as the first number of the sequence.

So you would do 1000(1001)/2 which is 500500.

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- 1 decade ago
1000²-(1000-1)²

+

998²-(998-1)²

+

...

= sum (j=1 to 500) [(2j)²-(2j-1)²]

= sum (j=1 to 500) [ 4j² - (4j² -4j +1)]

= sum (j=1 to 500) [4j - 1]

= 4*sum(j=1 to 500) j - sum(j=1 to 500) 1

= 4*(500²+500)/2 -500

= 2*(500²) + 2*500 - 500

= 2*(250000) + 500

= 500000 + 500

= 500500

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- Anonymous1 decade ago
program add;

var

sum:real;

a:integer;

begin

sum:=0;

for a:=1000 downto 0 do

begin

if a mod 2 = 0 then

sum:=sum+(a*a)

else

sum:=sum-(a*a);

end;

write(sum:0:0);

end.

wrote this program to solve it for you, output was 41,748. dont hold me responsible if its wrong tho lol

- Anonymous1 decade ago
555