anonymous
  • anonymous
the sum of the series 20c0 - 20c1 + 20c2 - 20c3+........+20c10 is
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
20(c0-c1+c2-c3 + ... + c10); what are the coeefs?
anonymous
  • anonymous
i think it should use binomial form (1-x)^20 or somewhat like it...
anonymous
  • anonymous
no no no hold on.. its a combinatorics problem. the 'c' here means that

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anonymous
  • anonymous
use (1-x)^20+(1+x)^20 you will get the ans...
anonymous
  • anonymous
deeprony have u understood the p problem??????
amistre64
  • amistre64
20!/20! - 20!/19! + 20!/18!2! .... like that?
anonymous
  • anonymous
"(1-x)^20+(1+x)^20" what is x here?? yea i understood that quetion=)
anonymous
  • anonymous
yes mistre thats correct
amistre64
  • amistre64
i stil gotta do these the brute force way ;)
anonymous
  • anonymous
isint there a formula??
anonymous
  • anonymous
(1-x)^20=20c0-20c1x+.......... now 20cn=20c20-n. and thus 20c20=20c0. and so on put x=1. 20c0-20c1+.....=0 2(20c0-20c1-.......+20c10)-20c10=0 hence ans is (20c10)/2
anonymous
  • anonymous
can you explain what you did hre.. 20c0-20c1+.....=0 2(20c0-20c1-.......+20c10)-20c10=0
anonymous
  • anonymous
i put just x=1 in the 1st expansion of (1-x)^20
anonymous
  • anonymous
yea ok wat did u do thn??
anonymous
  • anonymous
that is the way of doing.....using a binomial series....
anonymous
  • anonymous
i had (1-x)^20=20c0-20c1x+20c2x^2.... put x=1 0=20c0-20c1+20c2......
anonymous
  • anonymous
use 20c20=20c0,20c19=20c1,.......20c11=20c9.
anonymous
  • anonymous
i got that olryt... i didnt understnd what u did xt
anonymous
  • anonymous
*nxt
anonymous
  • anonymous
you get the term 20c0,20c1,....20c9 twice but 20c10 only one...
anonymous
  • anonymous
ahh i c... got it now...=))
anonymous
  • anonymous
thus i write .. 0=2(20c0-20c1+20c2........+20c10)-20c10
anonymous
  • anonymous
okkkkkkkkkkkkkkkkkk?????????
anonymous
  • anonymous
yea!
anonymous
  • anonymous
these are some tricks to do it...............
anonymous
  • anonymous
is the answer 0

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