binomial expansion help?

• Nov 11th 2009, 01:37 PM
Quacky
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 $|x|<4$, find in ascending powers upto and including the term $x^3$, the series expansion of
a) $(4-x)^1/2$
b) $(1+2x)(4-x)^1/2$

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

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

Thanks in advance for any replies.
• Nov 11th 2009, 05:32 PM
pencil09
Quote:

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 (Surprised):

Finally,
5) When $(1+ax)^n$ is expanded as a series in asceding powers of x, the co-efficients of x and $x^2$are -6 and 27 respectively. Find the values of a and n, the coefficient of $x^3$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,
$(1+ax)^n=nC0 (1)^n \left(ax\right)^0+ nC1 (1)^{(n-1)} (ax)^1+nC2 (1)^{(n-2)} (ax)^2+.....$ Note: $nCr=\frac {n!}{(n-r)! r!}$

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

(the co-efficients of $x^2$ and 27 )
$nC2 (1)^{(n-2)} (ax)^2=27$
$\frac {n!}{(n-2)! (2!)} (1) (a^2) (x^2) =27 (x^2)$ (cancel x^2)
$\frac {n (n-1) (n-2)!}{(n-2)! (2) (1)} (a^2) = 27$ (cancel (n-2)!)
$\frac {n (n-1)}{2} a^2 = 27$
$(n^2-n)(a^2)=54$........(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 $x^3$ easily, after u find the value of a and n....

....hope it help....(Happy)
• Nov 15th 2009, 10:21 AM
Quacky
Thanks, I'll have another go at the other 4 questions now.