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anonymous
 4 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... ?
anonymous
 4 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... ?

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

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.21+2+3+ .... + 2001 ... i guess so

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[2001^2−2000^2+1999^2−1998^2+...+3^2−2^2+1^2?\] this should be better.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(ab)^2= a^22ab+b^2 not a^2b^2 !!! O_O a^2b^2=(ab)(a+b)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Kreshnik fail me :))

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@ anonymoustwo44 it's ok ;)

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0or you can rewrite that as \[\sum_{i=1}^{n}(1)^{i+1}i^{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0mr. 123+1234 is correct

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

experimentX
 4 years ago
Best 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

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.21+2+3+ ... + 2001 ... regular ap

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

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2since 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
 4 years ago
Best ResponseYou've already chosen the best response.0OR 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@experiment44 that's just a formula for generating the series above not for solving its sum

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

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