Binomial theorem

[1141]

Can anyone help me?

I've just started doing the Binomial theorem, and I'm stuck on the question.
There are two parts to the question, part one is:

"Expand (1+2x)[sup:24dlw0vx]16[/sup:24dlw0vx] up to and including the term in x[sup:24dlw0vx]3[/sup:24dlw0vx]."

This part I understand fine. I can do this. The answer I got is:

1+32x+480x[sup:24dlw0vx]2[/sup:24dlw0vx]+4480x[sup:24dlw0vx]3[/sup:24dlw0vx]

I checked the answer in the book and it is correct.

It's the second part I do not understand:

"Deduce the coefficient of x[sup:24dlw0vx]3[/sup:24dlw0vx] in the expansion of (1+3x)(1+2x)[sup:24dlw0vx]16[/sup:24dlw0vx]."

I know the formula to find the coefficient, but I'm not sure how I would apply it in this question???

 

[soroban]

Hello, 1141!

Expand \(\displaystyle (1+2x)^{16}\) up to and including the term in \(\displaystyle x^3\)

The answer I got is: .\(\displaystyle 1+32x+480x^2+4480x^3\)

I checked the answer in the book and it is correct.

It's the second part I do not understand:

. . Deduce the coefficient of \(\displaystyle x^3\) in the expansion of \(\displaystyle (1+3x)(1+2x)^{16}\)

You've already done the hard part . . . now just multiply!


You have:. . \(\displaystyle (1+3x)(1+2x)^{16}\)

. . = . . . \(\displaystyle =\;(1+3x)(1 + 32x + 480x^2 + 4480x^3 + \hdots)\)


Multiply: . \(\displaystyle 1 + 32x + 480x^2 + 4480x^3 + \hdots\)
. . . . . . . . . . \(\displaystyle 3x + \;96x^2\; + 1440x^3 + 13440x^4 + \hdots\)


\(\displaystyle \text{Combine: }\;1 + 35x + 576x^2 + \underbrace{5920}_{\text{There!}}x^3 + \hdots\)

 

[1141]

Oh! Okay, I understand that.

Can you help with this question, sort of stuck on it too:

"Given that the coefficient of x in the expansion of (1+ax)(1+5x)[sup:1an3f9zl]40[/sup:1an3f9zl] is 207, determine the value of a."

 

[soroban]

Hello again, 1141!

\(\displaystyle \text{Given that the coefficient of }x\text{ in the expansion of }(1+ax)(1+5x)^{40}\text{ is 207,}\)

. . \(\displaystyle \text{determine the value of }a.\)

\(\displaystyle \text{Expand: }\1+5x)^{40} \;=\;1^{40} + 40(1)^{39}(5x)^1 + {40\choose2}(1)^{38}(5x)^2 + \hdots \;=\;1 + 200x + 19,500x^2 + \hdots\)


\(\displaystyle \text{So, we have: }\1 + ax)(1 + 200x + 19,500x^2 + \hdots)\)


Multiply it out:

. . \(\displaystyle \begin{array}{ccc}(1+ax)(1+200x + 19,500x^2 + \hdots) &=& 1 + 200x + 19,500x^2 + \hdots \\ & & \quad\;\;\; ax \;\;+ \;\;\;200ax^2 + \hdots \end{array}\)


\(\displaystyle \text{And we have: }\:1 + \underbrace{(a+200)}_{\text{This is 207}}x + (200a + 19,\!500)x^2 + \hdots\)


. . \(\displaystyle \text{Therefore: }\;a + 200 \:=\:207 \quad\Rightarrow\quad \boxed{a \:=\:7}\)