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Math Help - Binomial Theorem expanded

  1. #1
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    Binomial Theorem expanded

    ok i just want someone to double check my work please. for some reason i think i have the wrong answer.

    (use the binomial theorem to expand and simplify this expression)

    (x-y)^8


    ok so i used pascal's triangle ...the pascal triangle in my book stops at the 7th row so i got this for my 8th row 1,8,28,56,70,56,28,8,1

    so this is what i got ...Please check my work.

    1x^8-8x^7y+28x^6y^2-56x^5y^3+70x^4y^4-56x^3y^5+28x^2y^6-8xy^7+1y^8

    ok i know this looks confusing but if you copy and paste it into online calculator it should put everything into place. Did i do this right???
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  2. #2
    Super Member Matt Westwood's Avatar
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    If I was marking I's say "no", because it says "use the binomial theorem", not "read the appropriate row of Pascal's Triangle from your textbook".

    I reckon what you ought to do would be to use the identity

    \binom n m = \frac {n!} {(n-m)! m!}

    where n = 8 and m is each of 0 to 8 in turn. The arithmetic is not hard, and the numbers should work out the same as what you got.
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  3. #3
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    Quote Originally Posted by Matt Westwood View Post
    If I was marking I's say "no", because it says "use the binomial theorem", not "read the appropriate row of Pascal's Triangle from your textbook".

    I reckon what you ought to do would be to use the identity

    \binom n m = \frac {n!} {(n-m)! m!}

    where n = 8 and m is each of 0 to 8 in turn. The arithmetic is not hard, and the numbers should work out the same as what you got.
    i dont understand what your trying to tell me. its asking me to expand (x-y)^8

    what you gave me is no where in my book? your technique probably works too but im just no familiar with it sorry.
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  4. #4
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    Quote Originally Posted by flexus View Post
    ok i just want someone to double check my work please. for some reason i think i have the wrong answer.

    (use the binomial theorem to expand and simplify this expression)

    (x-y)^8


    ok so i used pascal's triangle ...the pascal triangle in my book stops at the 7th row so i got this for my 8th row 1,8,28,56,70,56,28,8,1

    so this is what i got ...Please check my work.

    1x^8-8x^7y+28x^6y^2-56x^5y^3+70x^4y^4-56x^3y^5+28x^2y^6-8xy^7+1y^8

    ok i know this looks confusing but if you copy and paste it into online calculator it should put everything into place. Did i do this right???
    HI

    you are correct .

    Instead of using the pascal triangle to look for the coefficient , you can actually use the identity which matt has suggested .
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  5. #5
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    Quote Originally Posted by mathaddict View Post
    HI

    you are correct .

    Instead of using the pascal triangle to look for the coefficient , you can actually use the identity which matt has suggested .
    could you show me how to do that? if its not to much trouble that is? thank you.
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  6. #6
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    im sorry i still don't get it so did i do it right or wrong?? is my answer correct?
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  7. #7
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    Show you how to do what? Multiply and divide? Matt told you that the coefficient of x^ny^{8-n} is \frac{8!}{n!(8- n)!}. Calculate that for n= 0 to 8 (Actually, you only need to do it for n= 0 t0 4, after that you get duplicates). Do you know that n!= n(n-1)(n-2)\cdot\cdot\cdot (3)(2)(1)? So that The coefficient for x^2y^6 is \left(\begin{array}{cc}8 \\ 2\end{array}\right)= \frac{8!}{2! 6!}= \frac{40320}{(2)(720)}. You can simplfy that a lot by observing that \frac{8!}{2! 6!}= \frac{8(7)(6)(5)(4)(3)(2)(1)}{(2(1))(6(5)(4)(3)(2)  (1)} and canceling a lot- you eventually get 4(7)= 28, just as you have!
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  8. #8
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    im sorry but no one has told me if i did it right or if i got the right answer???
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  9. #9
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    Quote Originally Posted by flexus View Post
    im sorry but no one has told me if i did it right or if i got the right answer???
    The answer is correct. But note that when a question gives an instruction to solve it in a particular way, your solution must show that this instruction has been followed ....
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