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natasha.aries

HEY PLEASE HELP ME! :) Find the first four terms in the expansion of (x^1/2+1)^30

  • one year ago
  • one year ago

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  1. RadEn
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    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

    • one year ago
  2. natasha.aries
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    wait im confused do you think you can write it out on paper and attach it? im sorry and i understand if you cannot.

    • one year ago
  3. RadEn
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    do know about combination ?

    • one year ago
  4. natasha.aries
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    oh wait! i think i get it! but do i go all the way to thirty? thats whats confusing me?

    • one year ago
  5. natasha.aries
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    no...:/

    • one year ago
  6. RadEn
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    the general formula of combination is |dw:1365545267017:dw| do u know about factorial :)

    • one year ago
  7. natasha.aries
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    oh wait! i do know about it loll what would be n and what would be r?

    • one year ago
  8. RadEn
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    actually, n and r are natural number

    • one year ago
  9. natasha.aries
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    so how would i use the formula?

    • one year ago
  10. RadEn
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    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 ?

    • one year ago
  11. natasha.aries
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    yes!

    • one year ago
  12. RadEn
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    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 ?

    • one year ago
  13. natasha.aries
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    yes!

    • one year ago
  14. RadEn
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    so, what is the 30C1 ? :) try

    • one year ago
  15. natasha.aries
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    30

    • one year ago
  16. RadEn
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    cool :)

    • one year ago
  17. natasha.aries
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    do i go all the way down until i get to 1?

    • one year ago
  18. RadEn
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    so, so on u can calculation of 30C2 and 30C3 right

    • one year ago
  19. natasha.aries
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    yeah

    • one year ago
  20. RadEn
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    ok, now we have the other problem about exponent

    • one year ago
  21. natasha.aries
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    ok

    • one year ago
  22. RadEn
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    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 ?

    • one year ago
  23. natasha.aries
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    yess

    • one year ago
  24. RadEn
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    so, u can miss them

    • one year ago
  25. RadEn
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    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 ?

    • one year ago
  26. RadEn
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    opp, x^15 i mean :)

    • one year ago
  27. natasha.aries
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    yes

    • one year ago
  28. RadEn
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    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)

    • one year ago
  29. RadEn
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    it just simplification in radical form to exponential form

    • one year ago
  30. RadEn
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    or just one expression sqrt(x^29) only

    • one year ago
  31. RadEn
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    now, what about u with (x^1/2)^(30-2) (1)^2 = ... ?

    • one year ago
  32. natasha.aries
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    ohh!! okay i understand! thank you very much!!!!

    • one year ago
  33. RadEn
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    you're welcome

    • one year ago
  34. natasha.aries
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    :)

    • one year ago
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