binomial series expansion

ive done part a and part b however the way I did part b was subbing the powers of x of the terms I calculated in part a and equallised it with its respective coefficient that I calculated in part a when expanding the binomial series. I know its not a safe way to determine the constant term and plus it didnt show how it could be expressed in that form. Is there a method of finding this expression? Or do you just have to spot a pattern by trial and error?

lastly im not sure how to tackle part c either.

thanks for any help in advanced

Posted from TSR Mobile

ive done part a and part b however the way I did part b was subbing the powers of x of the terms I calculated in part a and equallised it with its respective coefficient that I calculated in part a when expanding the binomial series. I know its not a safe way to determine the constant term and plus it didnt show how it could be expressed in that form. Is there a method of finding this expression? Or do you just have to spot a pattern by trial and error?

lastly im not sure how to tackle part c either.

thanks for any help in advanced

Posted from TSR Mobile

From what you've described in b) it seems fine, though I imagine you'd only need to do it for 1 value of n and check it for other values if you wish.

For c, what value of x do you need to make the thing before you expanded it look like (1/0.999)^3?

I got 0.01 but thats by equating (1/0.999)^3=(1-x/10)^-3.
I re arranged to get (1/0.999)^3 = 1/(1-x/10)^3 and cube rooted both sides and solved it algebraically from there which lead to 0.01?

That's the right value of x, and you got the same one both times. All good.

if i sub in the expanded form i get 0.9997... but the answer is like 1.00003...

Hmm, I'm not too sure then, sorry. You might want to follow what the other poster has suggested

for c)

$\displaystyle \left (\frac{1}{0.999}\right )^3=\left (\frac{1}{1-0.001}\right )^3=\left (1-10^{-3}\right )^{-3}$

For expansion use the following: (for 14 decimal places you need for first 5 terms, because (10^{-3})^5=10^(-15))

$\displaystyle \left (1-x\right )^{\alpha}=\sum_{k=0}^{5} \binom{\alpha}{k} (-1)^k\cdot x^k$

and

$\displaystyle \binom{\alpha}{k}=\frac{\alpha \cdot \left (\alpha - 1\right )\cdot ... \cdot \left (\alpha -k+1\right )}{k!}$

ie.

$\displaystyle \binom{-3}{2}=\frac{(-3)\cdot (-4)}{2\cdot 1}=\frac{12}{2}=6$