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

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  • 1 decade ago
    Favorite 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

  • 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
    1 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

  • 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.

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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

  • Anonymous
    4 years ago

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  • Anonymous
    1 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

  • Anonymous
    1 decade ago

    555

  • 1 decade ago

    The answer is

    500,500

    I love math...

    Any more???

  • MsMath
    Lv 7
    1 decade ago

    500,500

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