butterflydreamer
  • butterflydreamer
help please :) - BINOMIAL THEOREM CC's question below:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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butterflydreamer
  • butterflydreamer
Prove: \[1 + \left(\begin{matrix}10 \\ 2\end{matrix}\right)3^{2} + \left(\begin{matrix}10 \\ 4\end{matrix}\right)3^{4} + \left(\begin{matrix}10 \\ 6\end{matrix}\right)3^{6} + \left(\begin{matrix}10 \\ 8\end{matrix}\right)3^{8} + 3^{10} = 2^9 (2^{10} +1)\]
butterflydreamer
  • butterflydreamer
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 -.-
amistre64
  • amistre64
``` \left(\begin{matrix}10 \\ 2\end{matrix}\right) ``` is better writen as ``` \binom{10}{2} ``` \[\binom{10}{2}\]

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butterflydreamer
  • butterflydreamer
oh, thanks for the tip LOL
amistre64
  • amistre64
(x-1)^(10) = cx^10 -cx^9 +cx^8 -cx^7.. right?
amistre64
  • amistre64
what is: (x+1)^(10) + (x-1)^(10)
amistre64
  • amistre64
\[2 (x^{10}+45 x^8+210 x^6+210 x^4+45 x^2+1)\]
butterflydreamer
  • butterflydreamer
ohhh true true . That makes sense! So that's how they got the even parts only.
amistre64
  • amistre64
im an idiot, so i just use whatever comes to mind :)
amistre64
  • amistre64
oh im not sure how THEY did it .. but this might be a start
butterflydreamer
  • butterflydreamer
yeess. LOL i can't believe i expanded wrong xD oh gosh. Okay sooo, what should we do next?
amistre64
  • amistre64
dunno offhand, i was just trying to get your thought to get rid of the odd parts
amistre64
  • amistre64
since LHS and RHS are finite numbers, cant we just let x=3 and work the math to prove?
amistre64
  • amistre64
(x+1)^(10) + (x-1)^(10) 4^(10) + 2^(10) 2^(10) 2^10 + 2^(10) 2^(10) [2^(10)+1]
amistre64
  • amistre64
and divide by 2, since we are twice as heavy
amistre64
  • amistre64
or something like that
butterflydreamer
  • butterflydreamer
.... oh... o.o
butterflydreamer
  • butterflydreamer
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
amistre64
  • amistre64
direct solutions tend to work out great to me ... \[1 + \left(\begin{matrix}10 \\ 2\end{matrix}\right)x^{2} + \left(\begin{matrix}10 \\ 4\end{matrix}\right)x^{4} + \left(\begin{matrix}10 \\ 6\end{matrix}\right)x^{6} + \left(\begin{matrix}10 \\ 8\end{matrix}\right)x^{8} + x^{10} =\frac{(x+1)^{10}+(x-1)^{10}}{2}\]
amistre64
  • amistre64
when x=3, ive already worked the numerator to 2^(10) [2^(10)+1] right?
butterflydreamer
  • butterflydreamer
yes
amistre64
  • amistre64
then it is proofed
butterflydreamer
  • butterflydreamer
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.
amistre64
  • amistre64
\[1 + \left(\begin{matrix}10 \\ 2\end{matrix}\right)3^{2} + \left(\begin{matrix}10 \\ 4\end{matrix}\right)3^{4} + \left(\begin{matrix}10 \\ 6\end{matrix}\right)3^{6} + \left(\begin{matrix}10 \\ 8\end{matrix}\right)3^{8} + 3^{10}\] \[\frac{(3+1)^{10}+(3-1)^{10}}{2}\] \[\frac{4^{10}+2^{10}}{2}\] \[\frac{2^{10}(2^{10}+1)}{2}\] \[{2^{9}(2^{10}+1)}\]
amistre64
  • amistre64
well, it was your way, i just built on it :)
butterflydreamer
  • butterflydreamer
aahh okay! Thanks again :D!

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