anonymous
  • anonymous
\[\\text{How can I find the sum of:} 2001^2-2000^2+1999^2-1998^2+...+3^2-2^2+1^2 ?\] How can I simplify this... ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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experimentX
  • experimentX
(2001-2000)(2001+2000)+(1999+1998) ... + 3 +2+1
experimentX
  • experimentX
1+2+3+ .... + 2001 ... i guess so
anonymous
  • anonymous
\[2001^2−2000^2+1999^2−1998^2+...+3^2−2^2+1^2?\] this should be better.

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More answers

anonymous
  • anonymous
I got it
lgbasallote
  • lgbasallote
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?
anonymous
  • anonymous
(a-b)^2= a^2-2ab+b^2 not a^2-b^2 !!! O_O a^2-b^2=(a-b)(a+b)
anonymous
  • anonymous
@Kreshnik fail me :))
anonymous
  • anonymous
@ anonymoustwo44 it's ok ;)
Mr.Math
  • Mr.Math
Use this\[\sum_{n=1}^k n^2=\frac{1}{6}k(k+1)(2k+1).\]
experimentX
  • experimentX
pattern should be (-1)^(n+1) n^2
anonymous
  • anonymous
or you can rewrite that as \[\sum_{i=1}^{n}(-1)^{i+1}i^{2}\]
anonymous
  • anonymous
try it
anonymous
  • anonymous
thank you
anonymous
  • anonymous
mr. 123+1234 is correct
lgbasallote
  • lgbasallote
@Mr.Math is that mathematical induction? i would love to learn that..
experimentX
  • experimentX
(2001-2000)(2001+2000)+(1999+1998)(1999-1998)+ ... + (3+2)(3-2)+1 1(2001+2000)+1(1999+1998)+ ...3+2+1
experimentX
  • experimentX
1+2+3+ ... + 2001 ... regular ap
Mr.Math
  • Mr.Math
Yes, Mathematical induction is the best way to show that the sum above is true. But it's not how it's derived.
experimentX
  • experimentX
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
anonymous
  • anonymous
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
anonymous
  • anonymous
@experiment44 that's just a formula for generating the series above not for solving its sum
experimentX
  • experimentX
of course ,,,, after you find ap, finding sum is easy as piece of cake
experimentX
  • experimentX
thanks for medal ... one for you .. for giving complete answer.

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