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butterflydreamer

  • one year ago

help please :) - BINOMIAL THEOREM CC's question below:

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  1. butterflydreamer
    • one year ago
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    Prove: \[1 + \left(\begin{matrix}10 \\ 2\end{matrix}\right)3^{2} + \left(\begin{matrix}10 \\ 4\end{matrix}\right)3^{4} + \left(\begin{matrix}10 \\ 6\end{matrix}\right)3^{6} + \left(\begin{matrix}10 \\ 8\end{matrix}\right)3^{8} + 3^{10} = 2^9 (2^{10} +1)\]

  2. butterflydreamer
    • one year ago
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    I'm considering: \[(1+x)^{10} = \left(\begin{matrix}10 \\ 0\end{matrix}\right) +\left(\begin{matrix}10 \\ 1\end{matrix}\right)x + \left(\begin{matrix}10 \\ 2\end{matrix}\right)x^2 + \left(\begin{matrix}10 \\ 3\end{matrix}\right)x^3 + ... + x^{10} \] But i have no idea where to go from here. If i look at the RHS, then i came up with something....?: \[2^9 ( 2^{10} + 1) \rightarrow 2^{n-1} (2^{n} +1) \] But i'm not how to get there :/ I also noticed on the LHS, the things are all even. Idk -.-

  3. amistre64
    • one year ago
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    ``` \left(\begin{matrix}10 \\ 2\end{matrix}\right) ``` is better writen as ``` \binom{10}{2} ``` \[\binom{10}{2}\]

  4. butterflydreamer
    • one year ago
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    oh, thanks for the tip LOL

  5. amistre64
    • one year ago
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    (x-1)^(10) = cx^10 -cx^9 +cx^8 -cx^7.. right?

  6. amistre64
    • one year ago
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    what is: (x+1)^(10) + (x-1)^(10)

  7. amistre64
    • one year ago
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    \[2 (x^{10}+45 x^8+210 x^6+210 x^4+45 x^2+1)\]

  8. butterflydreamer
    • one year ago
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    ohhh true true . That makes sense! So that's how they got the even parts only.

  9. amistre64
    • one year ago
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    im an idiot, so i just use whatever comes to mind :)

  10. amistre64
    • one year ago
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    oh im not sure how THEY did it .. but this might be a start

  11. butterflydreamer
    • one year ago
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    yeess. LOL i can't believe i expanded wrong xD oh gosh. Okay sooo, what should we do next?

  12. amistre64
    • one year ago
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    dunno offhand, i was just trying to get your thought to get rid of the odd parts

  13. amistre64
    • one year ago
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    since LHS and RHS are finite numbers, cant we just let x=3 and work the math to prove?

  14. amistre64
    • one year ago
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    (x+1)^(10) + (x-1)^(10) 4^(10) + 2^(10) 2^(10) 2^10 + 2^(10) 2^(10) [2^(10)+1]

  15. amistre64
    • one year ago
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    and divide by 2, since we are twice as heavy

  16. amistre64
    • one year ago
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    or something like that

  17. butterflydreamer
    • one year ago
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    .... oh... o.o

  18. butterflydreamer
    • one year ago
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    can we solve the question like that though? LOL. I've been trying to sub all kinds of weird numbers everywhere and trying to integrate things -.- I've been getting nowhere tbh LOl

  19. amistre64
    • one year ago
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    direct solutions tend to work out great to me ... \[1 + \left(\begin{matrix}10 \\ 2\end{matrix}\right)x^{2} + \left(\begin{matrix}10 \\ 4\end{matrix}\right)x^{4} + \left(\begin{matrix}10 \\ 6\end{matrix}\right)x^{6} + \left(\begin{matrix}10 \\ 8\end{matrix}\right)x^{8} + x^{10} =\frac{(x+1)^{10}+(x-1)^{10}}{2}\]

  20. amistre64
    • one year ago
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    when x=3, ive already worked the numerator to 2^(10) [2^(10)+1] right?

  21. butterflydreamer
    • one year ago
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    yes

  22. amistre64
    • one year ago
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    then it is proofed

  23. butterflydreamer
    • one year ago
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    hmm okay then. Thank you! It's just that for the prev. questions i've been working on, i've had to use a range of differentiation/integration and substitution methods so i thought this question would've involved similar steps but thanks again! Your method was way simpler.

  24. amistre64
    • one year ago
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    \[1 + \left(\begin{matrix}10 \\ 2\end{matrix}\right)3^{2} + \left(\begin{matrix}10 \\ 4\end{matrix}\right)3^{4} + \left(\begin{matrix}10 \\ 6\end{matrix}\right)3^{6} + \left(\begin{matrix}10 \\ 8\end{matrix}\right)3^{8} + 3^{10}\] \[\frac{(3+1)^{10}+(3-1)^{10}}{2}\] \[\frac{4^{10}+2^{10}}{2}\] \[\frac{2^{10}(2^{10}+1)}{2}\] \[{2^{9}(2^{10}+1)}\]

  25. amistre64
    • one year ago
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    well, it was your way, i just built on it :)

  26. butterflydreamer
    • one year ago
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    aahh okay! Thanks again :D!

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