Question

I need help with this practice problem for my calculus prep book

Answer 1

Given:

`\mleft(2x^2y^3-(1)/(4)y\mright)^9`

To find the 5th term:

Using the binomial expansion of

`(a+b)^n=\sum ^n_(i=0)\binom{n}{i}a^(\mleft(n-i\mright))b^i`

Put n=9 and i=4 (Fifth term of the series)

We get,

`\begin{gathered} \binom{9}{5}a^((9-4))b^4=(9!)/((9-5)!5!)a^5b^4 \\ =(9!)/(4!5!)a^5b^4 \\ =(6*7*8*9)/(4*3*2)a^5b^4 \\ =(7*2*9)a^5b^4 \\ =126a^5b^4\ldots\ldots\ldots(1) \end{gathered}`

Here, we have

`\begin{gathered} a=2x^2y^3 \\ b=-(1)/(4)y \end{gathered}`

Substituting these values in (1) we get,

`\begin{gathered} 126(2x^2y^3)^5(-(1)/(4)y)^4=126(2^5)x^(10)y^1^5((1)/(4))^4y^4 \\ =(126*32)/(256)x^(10)y^(15)y^4 \\ =15.75x^(10)y^(19) \end{gathered}`

Hence, the fifth term of the expansion is,

`15.75x^(10)y^(19)`