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natasha.aries

  • one year ago

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

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  1. RadEn
    • one year ago
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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

  2. natasha.aries
    • one year ago
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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.

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

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

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

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

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

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

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

  10. RadEn
    • one year ago
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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 ?

  11. natasha.aries
    • one year ago
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    yes!

  12. RadEn
    • one year ago
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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 ?

  13. natasha.aries
    • one year ago
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    yes!

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

  15. natasha.aries
    • one year ago
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    30

  16. RadEn
    • one year ago
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    cool :)

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

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

  19. natasha.aries
    • one year ago
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    yeah

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

  21. natasha.aries
    • one year ago
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    ok

  22. RadEn
    • one year ago
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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 ?

  23. natasha.aries
    • one year ago
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    yess

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

  25. RadEn
    • one year ago
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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 ?

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

  27. natasha.aries
    • one year ago
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    yes

  28. RadEn
    • one year ago
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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)

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

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

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

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

  33. RadEn
    • one year ago
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    you're welcome

  34. natasha.aries
    • one year ago
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    :)

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