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anonymous

  • one year ago

Yet another problem Dx

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  1. anonymous
    • one year ago
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    Identify the first and last term of the binomial expansion for the problem below. (3 + 2a)5 = _____ + … + _____

  2. anonymous
    • one year ago
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    ok wait a min this is some long stuff

  3. anonymous
    • one year ago
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    I hate bionomials passionately :/

  4. anonymous
    • one year ago
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    Lol xD spelt that wrong

  5. anonymous
    • one year ago
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    \[(3+2a)^5=\left(\begin{matrix}6 \\ 0\end{matrix}\right)(2a)^6+\left(\begin{matrix}6 \\ 1\end{matrix}\right)(2a)^5(3)^1+\left(\begin{matrix}6 \\ 2\end{matrix}\right)(2a)^4(3)^2+\left(\begin{matrix}6 \\ 3\end{matrix}\right)(2a)^3(3)^3+\left(\begin{matrix}6 \\ 4\end{matrix}\right)(2a)^2(3)^4\]\[+\left(\begin{matrix}6 \\ 5\end{matrix}\right)(2a)(3)^5+\left(\begin{matrix}6 \\ 0\end{matrix}\right)(3)^6\]

  6. anonymous
    • one year ago
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    woah.... o.e what...?

  7. anonymous
    • one year ago
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    wait i made some mistake i did 6 instead of 5

  8. anonymous
    • one year ago
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    lemme do it again

  9. anonymous
    • one year ago
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    Alright, hey thank you for taking time to help me

  10. anonymous
    • one year ago
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    \[(3+2a)^5=\left(\begin{matrix}5 \\ 0\end{matrix}\right)(2a)^5+\left(\begin{matrix}5 \\ 1\end{matrix}\right)(2a)^4(3)+\left(\begin{matrix}5 \\ 2\end{matrix}\right)(2a)^3(3)^2+\left(\begin{matrix}5 \\ 3\end{matrix}\right)(2a)^2(3)^3\]\[+\left(\begin{matrix}5 \\ 4\end{matrix}\right)(2a)^1(3)^4+\left(\begin{matrix}5 \\ 5\end{matrix}\right)(3)^5\]

  11. anonymous
    • one year ago
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    I think I can comprehend that.

  12. anonymous
    • one year ago
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    \[\left(\begin{matrix}5 \\ 0\end{matrix}\right)\]means 5 choose 0 or nCr on your calculator

  13. anonymous
    • one year ago
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    Alrighty sounds simple enough.

  14. anonymous
    • one year ago
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    lets see what you get for your answer

  15. anonymous
    • one year ago
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    243+810a+1080a2+720a3+240a4+32a5 this is what I got for the expansion. Is this correct?

  16. anonymous
    • one year ago
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    yes and now ill spoil the fun for ya...theres an easy way

  17. anonymous
    • one year ago
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    since it asked for 1st and last term only

  18. anonymous
    • one year ago
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    you can actually just 2a^5 and 3^5 to get the answer

  19. anonymous
    • one year ago
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    Seriously?

  20. anonymous
    • one year ago
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    yea nCr of first and last time is always nC0 and nCn which is 1

  21. anonymous
    • one year ago
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    and since the power has to add up to n, and 1 of the term has power 0, its pretty safe to say you can do that

  22. anonymous
    • one year ago
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    \[(2a)^5=2^5a^5=32a^5\]\[3^5=243\]

  23. anonymous
    • one year ago
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    makes sense to you?

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