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Kreshnik

  • 3 years ago

\[\\text{How can I find the sum of:} 2001^2-2000^2+1999^2-1998^2+...+3^2-2^2+1^2 ?\] How can I simplify this... ?

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  1. experimentX
    • 3 years ago
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    (2001-2000)(2001+2000)+(1999+1998) ... + 3 +2+1

  2. experimentX
    • 3 years ago
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    1+2+3+ .... + 2001 ... i guess so

  3. Kreshnik
    • 3 years ago
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    \[2001^2−2000^2+1999^2−1998^2+...+3^2−2^2+1^2?\] this should be better.

  4. anonymoustwo44
    • 3 years ago
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    I got it

  5. lgbasallote
    • 3 years ago
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    it looks like the pattern is x^2 - (x-1)^2 - (x-2)^2 and so on...any ideas on the sum of that experimentx?

  6. Kreshnik
    • 3 years ago
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    (a-b)^2= a^2-2ab+b^2 not a^2-b^2 !!! O_O a^2-b^2=(a-b)(a+b)

  7. anonymoustwo44
    • 3 years ago
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    @Kreshnik fail me :))

  8. Kreshnik
    • 3 years ago
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    @ anonymoustwo44 it's ok ;)

  9. Mr.Math
    • 3 years ago
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    Use this\[\sum_{n=1}^k n^2=\frac{1}{6}k(k+1)(2k+1).\]

  10. experimentX
    • 3 years ago
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    pattern should be (-1)^(n+1) n^2

  11. anonymoustwo44
    • 3 years ago
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    or you can rewrite that as \[\sum_{i=1}^{n}(-1)^{i+1}i^{2}\]

  12. anonymoustwo44
    • 3 years ago
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    try it

  13. Kreshnik
    • 3 years ago
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    thank you

  14. Rohangrr
    • 3 years ago
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    mr. 123+1234 is correct

  15. lgbasallote
    • 3 years ago
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    @Mr.Math is that mathematical induction? i would love to learn that..

  16. experimentX
    • 3 years ago
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    (2001-2000)(2001+2000)+(1999+1998)(1999-1998)+ ... + (3+2)(3-2)+1 1(2001+2000)+1(1999+1998)+ ...3+2+1

  17. experimentX
    • 3 years ago
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    1+2+3+ ... + 2001 ... regular ap

  18. Mr.Math
    • 3 years ago
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    Yes, Mathematical induction is the best way to show that the sum above is true. But it's not how it's derived.

  19. experimentX
    • 3 years ago
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    since this is an alternating series, the above formula you mentioned does not work, the method must be similar to AP ... depending on the last term

  20. anonymoustwo44
    • 3 years ago
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    OR since a^2-b^2=(a+b)(a-b) then that could be written as (2001+2000)(2001-2000)+(1999+1998)(1999-1998)+...+(3+2)(3-2)+1 =(2001+2000)(1)+(1999+1998)(1)+...+(3+2)(1)+1 =2001+2000+1999+1998+...+3+2+1 =1+2+3+.....+1998+1999+2000+2001 this is an arithmetic series now we can use the formula \[s=(n/2)(a_{1}+a_{n})\] where a sub n is the last term of the series and and a sub 1 is the first term of the series and n is the number of terms in the series so let's compute for the sum... s=(2001/2)(1+2001) s=(2001/2)(2002) s=(2001)(1001) s=2003001 is the sum :D

  21. anonymoustwo44
    • 3 years ago
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    @experiment44 that's just a formula for generating the series above not for solving its sum

  22. experimentX
    • 3 years ago
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    of course ,,,, after you find ap, finding sum is easy as piece of cake

  23. experimentX
    • 3 years ago
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    thanks for medal ... one for you .. for giving complete answer.

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