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whats the value of this sigma :
1 ² - 2 ² + 3 ² - 4 ² + 5 ² - 6 ² + ... +199 ²
i really worked trying to solve it ... but all the doors were closed i tried to divide it into two and get they sigma of each one so the total is the answer but no way ...
who can help me or at least give me the idea and i'll be grateful
Last edited by luca-deltodesco (2007-02-27 07:38:18)
The Beginning Of All Things To End.
The End Of All Things To Come.
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luca-deltodesco thanks but i didnt get it ... can you explain more ?! with number if you dont mind
jane dont forget that we have positive and negative numbers so ...
thanks luca i think i got it ! thanks alot
i got the way but how to do it ?! sorry
maybe you can split that sum into two sums: 'i' odd and 'i' even. Then you just have to sum two expressions not involving (-1)^i-1
i thought about it but i got some difficulity to write that in sigma ... can you help ?!
maybe you can split that sum into two sums: 'i' odd and 'i' even. Then you just have to sum two expressions not involving (-1)^i-1
Like sigma i=1 to 99 of some function of 2i plus i from 1 to 100 of some function of 2i+1 (possibly reversed I can't care enough to think about it)
Nah, delta's way is far superior.
But change i-1 for i+1....for the sake of sake's sake.
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aha thanks sekky for saying that but its really hard for me to get out that in this way :
199
∑ = ( -1 ) ^ ( i + 1 ) × i ²
i=1
could you help me please ?!
Pairing up terms might well help you. (n+1)² - n² = 2n+1, so by pairing up consecutive terms (you can pair the first one with 0² without changing the result) you can simplify the summation quite a bit.
Why did the vector cross the road?
It wanted to be normal.
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you can write it like this :
(1 ² - 2 ²) + (3 ² - 4 ² ) + ( 5 ² - 6 ²) + ... +199 ²
-3 + - 7 + - 11 + ...
its a series now
or you can do it with sigma like this :
ImPo$$!BLe = NoTH!nG
Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...
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