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Mathematics 10 Online
OpenStudy (anonymous):

Yet another problem Dx

OpenStudy (anonymous):

Identify the first and last term of the binomial expansion for the problem below. (3 + 2a)5 = _____ + … + _____

OpenStudy (anonymous):

ok wait a min this is some long stuff

OpenStudy (anonymous):

I hate bionomials passionately :/

OpenStudy (anonymous):

Lol xD spelt that wrong

OpenStudy (anonymous):

\[(3+2a)^5=\left(\begin{matrix}6 \\ 0\end{matrix}\right)(2a)^6+\left(\begin{matrix}6 \\ 1\end{matrix}\right)(2a)^5(3)^1+\left(\begin{matrix}6 \\ 2\end{matrix}\right)(2a)^4(3)^2+\left(\begin{matrix}6 \\ 3\end{matrix}\right)(2a)^3(3)^3+\left(\begin{matrix}6 \\ 4\end{matrix}\right)(2a)^2(3)^4\]\[+\left(\begin{matrix}6 \\ 5\end{matrix}\right)(2a)(3)^5+\left(\begin{matrix}6 \\ 0\end{matrix}\right)(3)^6\]

OpenStudy (anonymous):

woah.... o.e what...?

OpenStudy (anonymous):

wait i made some mistake i did 6 instead of 5

OpenStudy (anonymous):

lemme do it again

OpenStudy (anonymous):

Alright, hey thank you for taking time to help me

OpenStudy (anonymous):

\[(3+2a)^5=\left(\begin{matrix}5 \\ 0\end{matrix}\right)(2a)^5+\left(\begin{matrix}5 \\ 1\end{matrix}\right)(2a)^4(3)+\left(\begin{matrix}5 \\ 2\end{matrix}\right)(2a)^3(3)^2+\left(\begin{matrix}5 \\ 3\end{matrix}\right)(2a)^2(3)^3\]\[+\left(\begin{matrix}5 \\ 4\end{matrix}\right)(2a)^1(3)^4+\left(\begin{matrix}5 \\ 5\end{matrix}\right)(3)^5\]

OpenStudy (anonymous):

I think I can comprehend that.

OpenStudy (anonymous):

\[\left(\begin{matrix}5 \\ 0\end{matrix}\right)\]means 5 choose 0 or nCr on your calculator

OpenStudy (anonymous):

Alrighty sounds simple enough.

OpenStudy (anonymous):

lets see what you get for your answer

OpenStudy (anonymous):

243+810a+1080a2+720a3+240a4+32a5 this is what I got for the expansion. Is this correct?

OpenStudy (anonymous):

yes and now ill spoil the fun for ya...theres an easy way

OpenStudy (anonymous):

since it asked for 1st and last term only

OpenStudy (anonymous):

you can actually just 2a^5 and 3^5 to get the answer

OpenStudy (anonymous):

Seriously?

OpenStudy (anonymous):

yea nCr of first and last time is always nC0 and nCn which is 1

OpenStudy (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

OpenStudy (anonymous):

\[(2a)^5=2^5a^5=32a^5\]\[3^5=243\]

OpenStudy (anonymous):

makes sense to you?

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