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I have tried to solve few number series problems but this one is little difficult for me to solve. kindly help me to solve this.

-1,35,143,323,?

Thanks in advance.

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  • $\begingroup$ Hm.. Maybe the first number should be $6$ instead of $-1$? $\endgroup$ – athin Mar 21 at 13:47
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    $\begingroup$ This looks like this puzzle may have been taken from somewhere (v.s. one you created) - if so please provide attribution to the source. And welcome to Puzzling SE! $\endgroup$ – PartyHatPanda Mar 21 at 14:05
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I think I got this:

0*0 -1 = -1
6*6 -1 = 35
12*12-1 = 143
18*18 -1 = 323
24*24 -1 = 575

or:

(6*0)^2 -1 = -1
(6*1)^2 -1 = 35
(6*2)^2-1 = 143
(6*3)^2 -1 = 323
(6*4)^2 -1 = 575

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  • $\begingroup$ Or Maybe there is a Shorter way that im missing $\endgroup$ – Bedi Mar 21 at 13:51
  • $\begingroup$ My comment might be confusing after the edit so I removed it, but +1 from me now $\endgroup$ – PunPun1000 Mar 21 at 14:29
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    $\begingroup$ Or (-1)x1, 5x7, 11x13, 17x19, 23x25, ... $\endgroup$ – OHO Mar 21 at 16:44
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    $\begingroup$ @OHO That's neat. It holds true in math that $x^2 = (x + y)(x - y) + y^2$. In this instance $y$ is just 1 in every case; so it's the same equation, just a different representation. You may have known that already, just thought it was interesting. $\endgroup$ – Lord Farquaad Mar 21 at 20:53
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Bedi’s got the idea. Another way to write it would be:

(6n)^2 - 1 where n goes from 0 to 4

So, the next number is

(6*4)^2 - 1 = 575

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