anonymous
  • anonymous
binomial theorem: determine third term in expansion of (x^2 + 2x + 1 ) ^6
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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.Sam.
  • .Sam.
Try factoring x^2 + 2x + 1
anonymous
  • anonymous
i.e. write it as \(((x+1)^2)^6\)
anonymous
  • anonymous
giving you \[(x+1)^{12}\] and expand that one

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anonymous
  • anonymous
what about a permtatio/ combinatio formula?
anonymous
  • anonymous
like nCr or somethig like that..
anonymous
  • anonymous
yeah, something like that
anonymous
  • anonymous
can someone answer the question using nCr which prevents you from having to do the long factoring?
anonymous
  • anonymous
first term is \(x^{12}\) second term is \(12x^{11}\)
Luigi0210
  • Luigi0210
\[\left(\begin{matrix}13 \\ 2\end{matrix}\right)x^{11}1^2\]
anonymous
  • anonymous
third term is \(\binom{12}{2}x^{10}\)
Luigi0210
  • Luigi0210
oops, that's suppose to be 12
anonymous
  • anonymous
and \(\binom{12}{2}\) is \(\frac{12\times 11}{2}=6\times 11=66\)
anonymous
  • anonymous
you might know \(\binom{12}{2}\) as \(_{12}C_2\) same thing
anonymous
  • anonymous
\[ncr=\frac{ n \left( n-1 \right)\left( n-2 \right)...\left( n-\left( r-1 \right) \right) }{ r*\left( r-1 \right)\left( r-2 \right)....3*2*1 }\]

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