Binomial expansion and differentiation

**Confuzzled?** · February 6th 2007, 08:08 AM

1.The first three terms of the expansion of (1+kx)^n are 1+6x+27x^2. Find k and n and hence find the term in x^3.

2.In the expansion of (1+ax)^n, in ascending powers of x, the coefficient of x is 12 and the coefficient of x^3 is six times the coefficient of x^2. Find a and n.

3.Find dy/dx in terms of x and y: 2sin2xcos3y=1

Could anyone help me with the above questions. Thanks

**CaptainBlack** · February 6th 2007, 10:52 AM

Originally Posted by Confuzzled?

1.The first three terms of the expansion of (1+kx)^n are 1+6x+27x^2. Find k and n and hence find the term in x^3.

$ (1+kx)^n=1+n(kx)+\frac{n(n-1)}{2}(kx)^2+... $

So $n\,k=6$ , and $k^2\,(n^2-n)/2=27$ .

Taking the second of these, we may rewite it as:

$((n\,k)^2-(nk)\,k)/2=(36-6k)/2=27$ , so $k=-3$ , and so $n=-2$ .

RonL

**CaptainBlack** · February 6th 2007, 11:02 AM

Originally Posted by Confuzzled?

1.The first three terms of the expansion of (1+kx)^n are 1+6x+27x^2. Find k and n and hence find the term in x^3.

3.Find dy/dx in terms of x and y: 2sin2xcos3y=1

if:

$2 \sin(2x) \cos(3y)=1$ ,

then:

$ \frac{d}{dx} [\sin(2x) \cos(3y)]=0 $

Then using the product rule:

$ 2 \cos(2x) \cos(3y) - 3 \sin(2x) \sin(3y) \frac{dy}{dx}=0 $

so rearranging gives:

$\frac{dy}{dx}=\frac{2 \cos(2x) \cos(3y)}{3 \sin(2x) \sin(3y) }=(2/3) \cot(2x) \cot(3y)$

RonL

**Soroban** · February 6th 2007, 11:02 AM

Hello, Confuzzled?!

Here's #3 . . .

3. Find $\frac{dy}{dx}$ in terms of $x$ and $y$ : . $2\sin2x\cos3y\:=\:1$

Differentiate implicitly: . $2\sin2x\cdot\left[\sin3y(-3)\frac{dy}{dx}\right] + 2\cdot\left[ 2\cos2x\right]\cdot\cos3y \:=\:0$

Then: . $\frac{dy}{dx}\:=\:\frac{-4\cos2x\cos3y}{-6\sin2x\sin3y} \:=\:\frac{2}{3}\!\cdot\!\frac{\cos2x}{\sin2x}\!\c dot\!\frac{\cos3y}{\sin3y} \:=\:\frac{2}{3}\cot2x\cot3y $

**Confuzzled?** · February 6th 2007, 11:27 AM

thank you guys for your help although I'm afraid to say that the answers you got for the first question Captain black are different to the answers that are in the back of the book.

**CaptainBlack** · February 6th 2007, 11:58 AM

Originally Posted by Confuzzled?

thank you guys for your help although I'm afraid to say that the answers you got for the first question Captain black are different to the answers that are in the back of the book.

Type now corrected

RonL

**Soroban** · February 6th 2007, 02:37 PM

Hello again, Confuzzled!

1.The first three terms of the expansion of $(1+kx)^n$ are: . $1+6x+27x^2$
Find $k$ and $n$ and hence find the coefficient of $x^3$

Binomial Theorem: . $(1 + kx)^n \;=\;\binom{n}{0}1^n + \binom{n}{1}(kx) + \binom{n}{2}(kx)^2 + \binom{n}{3}(kx)^3 + \cdots$

Hence, we have: .[1] $\binom{n}{1}k = 6$ .and [2] . $\binom{n}{2}k^2 = 27$

[1] becomes: . $nk \,=\,6$ [3]

[2] becomes: . $\frac{n(n-1)}{2}k^2 \,= \,27\quad\Rightarrow\quad n(n-1)k^2 \,=\,54$ [4]

Divide [4] by [3]: . $\frac{n(n-1)k^2}{nk} \:=\:\frac{54}{6}\quad\Rightarrow\quad(n-1)k \:=\:9$

Divide by [3]: . $\frac{(n-1)k}{nk} \:=\:\frac{9}{6}\quad\Rightarrow\quad\frac{n-1}{n} \,=\,\frac{3}{2}$

Solve for $n\!:\;\;2n - 2 \:=\:3n\quad\Rightarrow\quad\boxed{n = -2}$

Substitute into [3]: . $-2k \,=\,6\quad\Rightarrow\quad\boxed{k \,=\,-3}$

Then: . $\binom{n}{3}k^3 \:=\:\frac{(-2)(-3)(-4)}{6}(-3)^3\:=\:\boxed{108}$ . . . coefficient of $x^3$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Check

$(1 - 3x)^{-2} \;=\;\frac{1}{1 - 6x + 9x^2} \;=\;1 + 6x + 27x^2 + 108x^3 + \cdots$

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