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anonymous
 one year ago
Yet another problem Dx
anonymous
 one year ago
Yet another problem Dx

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Identify the first and last term of the binomial expansion for the problem below. (3 + 2a)5 = _____ + … + _____

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok wait a min this is some long stuff

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I hate bionomials passionately :/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Lol xD spelt that wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(3+2a)^5=\left(\begin{matrix}6 \\ 0\end{matrix}\right)(2a)^6+\left(\begin{matrix}6 \\ 1\end{matrix}\right)(2a)^5(3)^1+\left(\begin{matrix}6 \\ 2\end{matrix}\right)(2a)^4(3)^2+\left(\begin{matrix}6 \\ 3\end{matrix}\right)(2a)^3(3)^3+\left(\begin{matrix}6 \\ 4\end{matrix}\right)(2a)^2(3)^4\]\[+\left(\begin{matrix}6 \\ 5\end{matrix}\right)(2a)(3)^5+\left(\begin{matrix}6 \\ 0\end{matrix}\right)(3)^6\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0woah.... o.e what...?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait i made some mistake i did 6 instead of 5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, hey thank you for taking time to help me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(3+2a)^5=\left(\begin{matrix}5 \\ 0\end{matrix}\right)(2a)^5+\left(\begin{matrix}5 \\ 1\end{matrix}\right)(2a)^4(3)+\left(\begin{matrix}5 \\ 2\end{matrix}\right)(2a)^3(3)^2+\left(\begin{matrix}5 \\ 3\end{matrix}\right)(2a)^2(3)^3\]\[+\left(\begin{matrix}5 \\ 4\end{matrix}\right)(2a)^1(3)^4+\left(\begin{matrix}5 \\ 5\end{matrix}\right)(3)^5\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think I can comprehend that.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\left(\begin{matrix}5 \\ 0\end{matrix}\right)\]means 5 choose 0 or nCr on your calculator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alrighty sounds simple enough.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lets see what you get for your answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0243+810a+1080a2+720a3+240a4+32a5 this is what I got for the expansion. Is this correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes and now ill spoil the fun for ya...theres an easy way

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since it asked for 1st and last term only

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can actually just 2a^5 and 3^5 to get the answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea nCr of first and last time is always nC0 and nCn which is 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and 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
 one year ago
Best ResponseYou've already chosen the best response.0\[(2a)^5=2^5a^5=32a^5\]\[3^5=243\]
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