Ask your own question, for FREE!
Mathematics 13 Online
OpenStudy (anonymous):

Binomial Distribution, anyone? :)

OpenStudy (anonymous):

Use the Binomial Theorem to find the binomial expansion of the expression. \[(d - 5)^{6}\]

OpenStudy (mathmate):

Are you familiar with C(n,r) ?

OpenStudy (anonymous):

I got the answer choices here.. A) \[d ^{6} + 6d ^{5} + 15d ^{4} + 20d ^{3} + 15d ^{2} + 6d + 1\] B) \[d ^{6} + 30d ^{5} + 375d ^{4} + 2500d ^{3} + 9375d ^{2} + 18750d + + 15625\] C) \[d ^{6} - 30d ^{5} + 375d ^{4} - 2500d ^{3} + 9375d ^{2} - 18750d + 15625\]

OpenStudy (mathmate):

\[(a+b)^n\ = \sum \left(\begin{matrix}n \\ i\end{matrix}\right) a^i \ b^{n-i}\]

OpenStudy (mathmate):

The first term is d^6. If you expand it, the second term is na^(n-1)b which in the given case, n=6. a=d b=-5 so the second term is 6d^5 (-5) = -30d^5 That good enough to make a choice?

OpenStudy (anonymous):

Answer C!! :)

OpenStudy (anonymous):

oooh makes sense that the + and - are alternating since the equation at the beginning has a minus sign...

OpenStudy (anonymous):

thank you for your explanation!

OpenStudy (mathmate):

Right, and the choice is correct!

OpenStudy (anonymous):

So if my equation is \[(d + 3)^{7} \] it has all + and no -. And the second term is also na^(n-1)b?

OpenStudy (mathmate):

Yep, would you like to give it a try?

OpenStudy (anonymous):

the second term is \[7d^{6}\] ?

OpenStudy (mathmate):

Almost, you just missed out the last bit!

OpenStudy (anonymous):

I missed the times 3...

OpenStudy (mathmate):

21d^6 it is! Good job!

OpenStudy (anonymous):

so it's \[21d ^{6}\] ?

OpenStudy (anonymous):

yees thank you!! :)

OpenStudy (mathmate):

No problem. If you like to try the third term, you're welcome to do so.

OpenStudy (anonymous):

i got it all right thanks to you! thank you so much. best response given :)

OpenStudy (mathmate):

You're welcome! :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!