1. ## binomial expansion help?

Ok I have a few questions here and I've tried to solve them all myself, could you please check?

2)Given that $\displaystyle |x|<4$, find in ascending powers upto and including the term $\displaystyle x^3$, the series expansion of
a) $\displaystyle (4-x)^1/2$
b) $\displaystyle (1+2x)(4-x)^1/2$

3)a) Find the first four terms of the expansion in ascending powers of x of $\displaystyle (2+3x)^-1, |x|<2/3$
b)Hence or otherwis, find the first four non-zero terms of the expansion, in ascending powers of x, of
$\displaystyle (1+x)(2+3x)^-1, |x|<2/3$

4)f$\displaystyle (x)=(1+3x)^-1, |x|<1/3$
a)Expand f(x) in ascending powers of x up to and including the term in $\displaystyle x^3$
b) Hence show for small x that
$\displaystyle (1+x)(1+3x)^-1 = 1-2x+6x^2-18x^3$

Thanks in advance for any replies.

2. Originally Posted by Quacky
I don't know if this is the correct thread as the titles can be so confusing, so I apologise If I've got this wrong.

Ok I have a few questions here and I've tried to solve them all myself unsuccessfully. I know it would take hours to finish them all, so could you at least provide me with a helpful tutorial as I'll have to go to bed now and these are due first thing tomorrow :

Finally,
5) When $\displaystyle (1+ax)^n$ is expanded as a series in asceding powers of x, the co-efficients of x and $\displaystyle x^2$are -6 and 27 respectively. Find the values of a and n, the coefficient of $\displaystyle x^3$and state the expressions for which the expansion is valid.

Thanks in advance for any replies.
i'll try the last one...
expand the equation,
$\displaystyle (1+ax)^n=nC0 (1)^n \left(ax\right)^0+ nC1 (1)^{(n-1)} (ax)^1+nC2 (1)^{(n-2)} (ax)^2+.....$ Note: $\displaystyle nCr=\frac {n!}{(n-r)! r!}$

(the co-efficients of x are -6)
$\displaystyle nC1 (1)^{(n-1)} (ax)^1=6x$ Note: $\displaystyle 1^n=1$
$\displaystyle \frac {n!}{(n-1)! 1!} (1) (a) (x)=(6) (x)$ (cancel x)
$\displaystyle \frac {n (n-1)!}{(n-1)!} (a) =6$ (cancel (n-1)!)
so,$\displaystyle (n) (a) = 6$........... (equation 1)

(the co-efficients of$\displaystyle x^2$ and 27 )
$\displaystyle nC2 (1)^{(n-2)} (ax)^2=27$
$\displaystyle \frac {n!}{(n-2)! (2!)} (1) (a^2) (x^2) =27 (x^2)$ (cancel x^2)
$\displaystyle \frac {n (n-1) (n-2)!}{(n-2)! (2) (1)} (a^2) = 27$ (cancel (n-2)!)
$\displaystyle \frac {n (n-1)}{2} a^2 = 27$
$\displaystyle (n^2-n)(a^2)=54$........(equation 2)

Use the substitution method to solve (equation 1) and (equation 2) and find the value of n and a....

You can find the coefficient for $\displaystyle x^3$ easily, after u find the value of a and n....

....hope it help....

3. Thanks, I'll have another go at the other 4 questions now.