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