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Yet another problem Dx

Mathematics
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Identify the first and last term of the binomial expansion for the problem below. (3 + 2a)5 = _____ + … + _____
ok wait a min this is some long stuff
I hate bionomials passionately :/

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

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