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\[\\text{How can I find the sum of:} 2001^22000^2+1999^21998^2+...+3^22^2+1^2 ?\]
How can I simplify this... ?
 2 years ago
 2 years ago
\[\\text{How can I find the sum of:} 2001^22000^2+1999^21998^2+...+3^22^2+1^2 ?\] How can I simplify this... ?
 2 years ago
 2 years ago

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experimentXBest ResponseYou've already chosen the best response.2
(20012000)(2001+2000)+(1999+1998) ... + 3 +2+1
 2 years ago

experimentXBest ResponseYou've already chosen the best response.2
1+2+3+ .... + 2001 ... i guess so
 2 years ago

KreshnikBest ResponseYou've already chosen the best response.1
\[2001^2−2000^2+1999^2−1998^2+...+3^2−2^2+1^2?\] this should be better.
 2 years ago

lgbasalloteBest ResponseYou've already chosen the best response.0
it looks like the pattern is x^2  (x1)^2  (x2)^2 and so on...any ideas on the sum of that experimentx?
 2 years ago

KreshnikBest ResponseYou've already chosen the best response.1
(ab)^2= a^22ab+b^2 not a^2b^2 !!! O_O a^2b^2=(ab)(a+b)
 2 years ago

anonymoustwo44Best ResponseYou've already chosen the best response.2
@Kreshnik fail me :))
 2 years ago

KreshnikBest ResponseYou've already chosen the best response.1
@ anonymoustwo44 it's ok ;)
 2 years ago

Mr.MathBest ResponseYou've already chosen the best response.0
Use this\[\sum_{n=1}^k n^2=\frac{1}{6}k(k+1)(2k+1).\]
 2 years ago

experimentXBest ResponseYou've already chosen the best response.2
pattern should be (1)^(n+1) n^2
 2 years ago

anonymoustwo44Best ResponseYou've already chosen the best response.2
or you can rewrite that as \[\sum_{i=1}^{n}(1)^{i+1}i^{2}\]
 2 years ago

RohangrrBest ResponseYou've already chosen the best response.0
mr. 123+1234 is correct
 2 years ago

lgbasalloteBest ResponseYou've already chosen the best response.0
@Mr.Math is that mathematical induction? i would love to learn that..
 2 years ago

experimentXBest ResponseYou've already chosen the best response.2
(20012000)(2001+2000)+(1999+1998)(19991998)+ ... + (3+2)(32)+1 1(2001+2000)+1(1999+1998)+ ...3+2+1
 2 years ago

experimentXBest ResponseYou've already chosen the best response.2
1+2+3+ ... + 2001 ... regular ap
 2 years ago

Mr.MathBest ResponseYou've already chosen the best response.0
Yes, Mathematical induction is the best way to show that the sum above is true. But it's not how it's derived.
 2 years ago

experimentXBest ResponseYou've already chosen the best response.2
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
 2 years ago

anonymoustwo44Best ResponseYou've already chosen the best response.2
OR since a^2b^2=(a+b)(ab) then that could be written as (2001+2000)(20012000)+(1999+1998)(19991998)+...+(3+2)(32)+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
 2 years ago

anonymoustwo44Best ResponseYou've already chosen the best response.2
@experiment44 that's just a formula for generating the series above not for solving its sum
 2 years ago

experimentXBest ResponseYou've already chosen the best response.2
of course ,,,, after you find ap, finding sum is easy as piece of cake
 2 years ago

experimentXBest ResponseYou've already chosen the best response.2
thanks for medal ... one for you .. for giving complete answer.
 2 years ago
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