anonymous
  • anonymous
HEY PLEASE HELP ME! :) Find the first four terms in the expansion of (x^1/2+1)^30
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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RadEn
  • RadEn
use this : 30C0 (x^1/2)^30(1)^0 + 30C1 (x^1/2)^(30-1) (1)^1 + 30C2 (x^1/2)^(30-2) (1)^2 + 30C3 (x^1/2)^(30-3) (1)^3 calculate and simplify it
anonymous
  • anonymous
wait im confused do you think you can write it out on paper and attach it? im sorry and i understand if you cannot.
RadEn
  • RadEn
do know about combination ?

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anonymous
  • anonymous
oh wait! i think i get it! but do i go all the way to thirty? thats whats confusing me?
anonymous
  • anonymous
no...:/
RadEn
  • RadEn
the general formula of combination is |dw:1365545267017:dw| do u know about factorial :)
anonymous
  • anonymous
oh wait! i do know about it loll what would be n and what would be r?
RadEn
  • RadEn
actually, n and r are natural number
anonymous
  • anonymous
so how would i use the formula?
RadEn
  • RadEn
well, first u have to know about factorial defined n! = n*(n-1)*(n-2)*... 3 . 2. 1 an example : 4! = 4 . 3 . 2 . 1 = 24 so far so good ?
anonymous
  • anonymous
yes!
RadEn
  • RadEn
ah, as extra info defined 0! = 1 nah, let's going to combination above 30C0 = 30!/(30-0)!0! = 30!/30!*1 = 30!/30! = 1, right ?
anonymous
  • anonymous
yes!
RadEn
  • RadEn
so, what is the 30C1 ? :) try
anonymous
  • anonymous
30
RadEn
  • RadEn
cool :)
anonymous
  • anonymous
do i go all the way down until i get to 1?
RadEn
  • RadEn
so, so on u can calculation of 30C2 and 30C3 right
anonymous
  • anonymous
yeah
RadEn
  • RadEn
ok, now we have the other problem about exponent
anonymous
  • anonymous
ok
RadEn
  • RadEn
look, that i wrote expression about 1^0, 1^1, 1^2, 1^3. these are be 1 right ? and whatever if multiplied by 1 be its self, right ?
anonymous
  • anonymous
yess
RadEn
  • RadEn
so, u can miss them
RadEn
  • RadEn
then for simplify of (x^1/2)^30, u can use the property of exponent : (a^m)^n = a^(m*n) so, (x^1/2)^30 = x^16, right ?
RadEn
  • RadEn
opp, x^15 i mean :)
anonymous
  • anonymous
yes
RadEn
  • RadEn
then so on, u have to simplify about (x^(1/2))^(30-1) = (x^(1/2))^(29) = x^(29/2) = sqrt(x^29) or u can modif again be sqrt(x^28 * x) = sqrt(x^28) sqrt(x) = (x^(28/2)) * sqrt(x) = x^14 * sqrt(x)
RadEn
  • RadEn
it just simplification in radical form to exponential form
RadEn
  • RadEn
or just one expression sqrt(x^29) only
RadEn
  • RadEn
now, what about u with (x^1/2)^(30-2) (1)^2 = ... ?
anonymous
  • anonymous
ohh!! okay i understand! thank you very much!!!!
RadEn
  • RadEn
you're welcome
anonymous
  • anonymous
:)

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