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anonymous

  • 5 years ago

find the coefficient of x^6 in the expansion of(5+2x^2)^7

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  1. anonymous
    • 5 years ago
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    my memory fails me on this - ill go and check

  2. anonymous
    • 5 years ago
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    It uses Mimi's Crazy theorem ;)

  3. Mimi_x3
    • 5 years ago
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    \[t_{k} = \left(\begin{matrix}n \\ k\end{matrix}\right)a^{n-k} b^{k} \]

  4. anonymous
    • 5 years ago
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    Mimi, that should be \( t_{k+1} \)

  5. Mimi_x3
    • 5 years ago
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    \[t_{6}= \left(\begin{matrix}7 \\ 6\end{matrix}\right)5\times(2x^2)^6\]

  6. Mimi_x3
    • 5 years ago
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    Nope..its only finding the coeffient

  7. Mimi_x3
    • 5 years ago
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    Not the independent term i think..

  8. anonymous
    • 5 years ago
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    You think wrong then :P

  9. anonymous
    • 5 years ago
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    its 7C3 5^4 2^3 = 35 * 625 * 8 = 175,000

  10. anonymous
    • 5 years ago
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    It's crazy right Mimi? lol

  11. anonymous
    • 5 years ago
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    where came from 7c3 5^4 2^3

  12. Mimi_x3
    • 5 years ago
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    I think that im not wrong, k+1 is used for something else..cant remember..

  13. Mimi_x3
    • 5 years ago
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    & this is not crazy, the other part is crazy ::

  14. anonymous
    • 5 years ago
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    because you have 2x^2 in the paraenthesis it will be the fourth term in the expansion so formula is 7C3 * 5^(7-3) *2^3

  15. anonymous
    • 5 years ago
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    what this teory name?

  16. Mimi_x3
    • 5 years ago
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    Binomial theorem.

  17. Mimi_x3
    • 5 years ago
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    wait..im wrong.

  18. anonymous
    • 5 years ago
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    i used the formula for the (r+1) th term of the binomial expansio (a+x)^n (r+1)th term = nC r a^(n-r) x^r

  19. anonymous
    • 5 years ago
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    Mimi, I am sure that the general term of \( (a+b)^n \) is \( t_{k+1} = \left(\begin{matrix}n \\ k\end{matrix}\right)a^{n-k} b^{k} \)

  20. anonymous
    • 5 years ago
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    yup - thats it Fool

  21. Mimi_x3
    • 5 years ago
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    Well, in my book it says.. "Specially for the expansion of (a+b)^n , the k+1th term is...that thing..cbb typing it again

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