mascga3 asked in Science & MathematicsMathematics · 1 decade ago

# 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

Relevance
• 1 decade ago

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

• 1 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.

• Anonymous

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

• 1 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.

• 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

• Anonymous
4 years ago

on the verge of the 2nd super melancholy; Iranian nukes, Iran(Hezbolah/Hamas)&Israeli conflict; India&Pakistani conflict; Russia&Georgia conflict; REX eighty 4, King Alfred Plan, FEMA coffins, FEMA camps, secret prisons, Patriot Act, tapping public telephones, analyzing public e-mails, RFID Verichips, XMARK, Mondex, SOMARK tattoos, Operation city Warrior, Operation unexpected impact, MJTF, undertaking Bluebeam, pretend ‘aliens’, LHC, Operation Mockingbird, schedule 21, NESARA, entomopters, DARPA monstrosities, Iron Mountain, Denver international Airport artwork of paintings, Georgia Guidestones, Bohemian Grove, authentic id, HR 1022, vile Bush govt orders(definite they deliver about over!) on the topic of martial regulation and putting human beings into artwork communities, warrantless searches, end of Habeas Corpus, and the upward push of a clean character spoke of as 'Homegrown Terrorists' as laid out in Senate invoice S.1959 which would be all people against the North American Union and the hot international Order enormously Christians! 2012 prophecies from each corner of the globe.

• Anonymous

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

• Anonymous

555

• 1 decade ago

500,500

I love math...

Any more???

• MsMath
Lv 7