# Thread: Binomial expansion and differentiation

1. ## Binomial expansion and differentiation

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

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

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

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

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

RonL

3. 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:

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

then:

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

Then using the product rule:

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

so rearranging gives:

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

RonL

4. Hello, Confuzzled?!

Here's #3 . . .

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

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

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

5. 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.

6. 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

7. Hello again, Confuzzled!

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

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

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

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

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

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

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

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

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

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

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Check

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

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# if the first three terms in the expansion of (1 kx)^n in the ascending power of x are 1-6x 33/2x^2 find the value of k and n

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