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(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.

I got it

(a-b)^2= a^2-2ab+b^2 not a^2-b^2 !!! O_O
a^2-b^2=(a-b)(a+b)

@ 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 \[\sum_{i=1}^{n}(-1)^{i+1}i^{2}\]

try it

thank you

mr. 123+1234 is correct

@Mr.Math is that mathematical induction? i would love to learn that..

1+2+3+ ... + 2001 ... regular ap

@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.