# Thread: Unusual Binomial Expansion Questions

1. ## Unusual Binomial Expansion Questions

I usually find Binomial Expansion questions quite straight forward but these ones have got me quite confused.

The first is made of two parts, the first being very simple, the second also seems simple but i'm not sure if i'm doing the correct thing.

1.
a) Expand $(1+3x)^-2$ in ascending powers of x up to and including the term in $x^3$, simplifying each term.

This part was simple and i believe the answer is $1 - 6x + 27x^2 - 108x^3 ...$

b) Hence, or otherwise, find the first three terms in the expansion of $\frac{x+4}{(1+3x)^2}$

For this i tried writing it out simply as $(x+4)(1+3x)^-2$ and then expanding the brackets using the answer from the first part to give a final answer of: $4 - 23x + 102x^2$ but this just seems too simple. Have I done it correctly or is there another method?

I have very little idea about the second question since i haven't encountered a similar one before:

2. When $(1 + ax)^n$ is expanded as a series in ascending powers of x, the coefficients of x and $x^2$ are -6 and 27 respectively.

a) Find the value of a and the value of n.

b) Find the coefficient of $x^3$.

c) State the set of values of x for which the expansion is valid.

I have attempted this question by trying to find simultaneous equations by comparing coefficients but I do not think it is the correct method since it seemed to be getting too complicated.

If someone could tell me whether i have done the first question correctly and show me how to do the second i would greatly appreciate it. Thanks in advance!

2. ## Binomial Coefficients

Hello Big_Joe
Originally Posted by Big_Joe
1.
a) Expand $(1+3x)^-2$ in ascending powers of x up to and including the term in $x^3$, simplifying each term.

This part was simple and i believe the answer is $1 - 6x + 27x^2 - 108x^3 ...$

b) Hence, or otherwise, find the first three terms in the expansion of $\frac{x+4}{(1+3x)^2}$

For this i tried writing it out simply as $(x+4)(1+3x)^-2$ and then expanding the brackets using the answer from the first part to give a final answer of: $4 - 23x + 102x^2$ but this just seems too simple. Have I done it correctly or is there another method?
Correct! Well done.

2. When $(1 + ax)^n$ is expanded as a series in ascending powers of x, the coefficients of x and $x^2$ are -6 and 27 respectively.

a) Find the value of a and the value of n.

b) Find the coefficient of $x^3$.

c) State the set of values of x for which the expansion is valid.

I have attempted this question by trying to find simultaneous equations by comparing coefficients but I do not think it is the correct method since it seemed to be getting too complicated.

If someone could tell me whether i have done the first question correctly and show me how to do the second i would greatly appreciate it. Thanks in advance!
Simultaneous equations is the way to go (unless, that is, that you happen to spot that it's the same as question 1!). So you get:

$(1+ax)^n = 1 + nax + \tfrac12n(n-1)a^2x^2 + \dots \equiv 1-6x+27x^2 + \dots$

$\Rightarrow na = -6 \Rightarrow a = -\frac6n$ (1)

and $\tfrac12n(n-1)a^2=27$ (2)

Substitute from (1) into (2):

$n(n-1)\frac{36}{n^2}= 54$

$\Rightarrow 36(n-1) = 54n$

$\Rightarrow n = -2$

$\Rightarrow a = 3$

So we have $(1+3x)^{-2} = 1 -6x + 27x^2 -108x^3 + \dots$

So the coefficient of $x^3$ is $-108$, and the expansion is valid provided $|3x| < 1$; i.e. $|x| < \tfrac13$