Question

I need some help please!
Use the Binomial Theorem to expand the binomial.

(3v + s)^5

a.) s5 – 5s4v + 10s3v2 – 10s2v3 + 5sv4 – v5
b.) s5 + 15s4v + 90s3v2 + 270s2v3 + 405sv4 + 243v5
c.) s5 + 45s4v + 270s3v2 + 810s2v3 + 1,215sv4 + 729v5
d.) s5 + 15s4 + 90s3 + 270s2 + 405s + 243

Answer 1

there is not a minus sign in sight for forget the first answer.

Answer 2

\((s+3v)^5\) first term is \(s^5\)
second term is \(5\times s^4\times 3v\) or \(15s^4v\) the 5 because \(_5C_1=5\)

Answer 3

that means C is out as well as A

Answer 4

D is also out because it doesn't even have a \(v\) in it. so i guess by process of elimination it must be B

Answer 5

Alright thanks :D

Answer 6

i also cant find the fourth term of (d – 4b)^3 :/ think you can help?

Answer 7

terms of \((a+b)^5\) look like
\[a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\] you can get the coefficients off of pascal's triangle

Answer 8

heres the possible choice for it
b3
–b3
64b3
–64b3

Answer 9

there are only 4 terms in \((a-4b)^3\) the last term is \((-4b)^3=-64b^3\)

Answer 10

alright, thanks a ton! :D

Answer 11

yw