TrojanPoem
  • TrojanPoem
Prove that: [(2n+1)C(0)]^2 - [(2n+1)C(1)]^2 + [(2n+1)C(2)]^2 -..... - ([(2n+1)C(2n+1)]^2 = 0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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TrojanPoem
  • TrojanPoem
Someone rewrite it with EQN I can't.
dan815
  • dan815
like thsi right?
dan815
  • dan815
|dw:1435164304284:dw|

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More answers

TrojanPoem
  • TrojanPoem
Use another method this E is out of my curriculum.
dan815
  • dan815
it just means add
dan815
  • dan815
but i think this ons is pretty straight forawrd because
dan815
  • dan815
we have symmetry along the pascal triangle
TrojanPoem
  • TrojanPoem
You know , it's meant to be proved with binomial
dan815
  • dan815
that is binomial
dan815
  • dan815
[(2n+1)C(0)]^2 = [(2n+1)C(2n+1)]^2 [(2n+1)C(1)]^2=[(2n+1)C(2n)]^2 [(2n+1)C(2)]^2=[(2n+1)C(2n-1)]^2 and so on right
dan815
  • dan815
im not satisfied with this though something is not making sense let me check something
dan815
  • dan815
|dw:1435164616647:dw|
dan815
  • dan815
ok silly yes it works xD

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