anonymous
  • anonymous
Yet another problem Dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Identify the first and last term of the binomial expansion for the problem below. (3 + 2a)5 = _____ + … + _____
anonymous
  • anonymous
ok wait a min this is some long stuff
anonymous
  • anonymous
I hate bionomials passionately :/

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anonymous
  • anonymous
Lol xD spelt that wrong
anonymous
  • anonymous
\[(3+2a)^5=\left(\begin{matrix}6 \\ 0\end{matrix}\right)(2a)^6+\left(\begin{matrix}6 \\ 1\end{matrix}\right)(2a)^5(3)^1+\left(\begin{matrix}6 \\ 2\end{matrix}\right)(2a)^4(3)^2+\left(\begin{matrix}6 \\ 3\end{matrix}\right)(2a)^3(3)^3+\left(\begin{matrix}6 \\ 4\end{matrix}\right)(2a)^2(3)^4\]\[+\left(\begin{matrix}6 \\ 5\end{matrix}\right)(2a)(3)^5+\left(\begin{matrix}6 \\ 0\end{matrix}\right)(3)^6\]
anonymous
  • anonymous
woah.... o.e what...?
anonymous
  • anonymous
wait i made some mistake i did 6 instead of 5
anonymous
  • anonymous
lemme do it again
anonymous
  • anonymous
Alright, hey thank you for taking time to help me
anonymous
  • anonymous
\[(3+2a)^5=\left(\begin{matrix}5 \\ 0\end{matrix}\right)(2a)^5+\left(\begin{matrix}5 \\ 1\end{matrix}\right)(2a)^4(3)+\left(\begin{matrix}5 \\ 2\end{matrix}\right)(2a)^3(3)^2+\left(\begin{matrix}5 \\ 3\end{matrix}\right)(2a)^2(3)^3\]\[+\left(\begin{matrix}5 \\ 4\end{matrix}\right)(2a)^1(3)^4+\left(\begin{matrix}5 \\ 5\end{matrix}\right)(3)^5\]
anonymous
  • anonymous
I think I can comprehend that.
anonymous
  • anonymous
\[\left(\begin{matrix}5 \\ 0\end{matrix}\right)\]means 5 choose 0 or nCr on your calculator
anonymous
  • anonymous
Alrighty sounds simple enough.
anonymous
  • anonymous
lets see what you get for your answer
anonymous
  • anonymous
243+810a+1080a2+720a3+240a4+32a5 this is what I got for the expansion. Is this correct?
anonymous
  • anonymous
yes and now ill spoil the fun for ya...theres an easy way
anonymous
  • anonymous
since it asked for 1st and last term only
anonymous
  • anonymous
you can actually just 2a^5 and 3^5 to get the answer
anonymous
  • anonymous
Seriously?
anonymous
  • anonymous
yea nCr of first and last time is always nC0 and nCn which is 1
anonymous
  • anonymous
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
anonymous
  • anonymous
\[(2a)^5=2^5a^5=32a^5\]\[3^5=243\]
anonymous
  • anonymous
makes sense to you?

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