Question

find the 5th term in the expansion of (3x^2-sqrt(y))^6

a)135x^4y^2
b)45x^4y^2
c)-135x^4y^2
d)-45x^4y^2

Answer 1

binomial theorem anyone?

Answer 2

i think its a)135x^4y^2 right?

Answer 3

exponent is 6
term number is 5

\[T_{5} = \left(\begin{matrix}6 \\ 5\end{matrix}\right)a^{6-5} b^{5}\]
where a = 3x^2
b = -sqrt(y)

Answer 4

wait i did something wrong there
the answer is A

Answer 5

it should be
\[\left(\begin{matrix}6 \\ 4\end{matrix}\right)(3x^{2})^{6-4} (-\sqrt{y})^{4} = 15*9*x^{4}*y^{2}\]