# Binomial expansion help- is the question wrong?

• Feb 28th 2010, 07:42 AM
fishkeeper
Binomial expansion help- is the question wrong?
Quote:

Find the coefficiant of $\displaystyle x$ in the expansion of $\displaystyle (1 - 2x)^4 (2 + x)^9$
Ive done this (I think) and got:

$\displaystyle 512 - 18432 x^2 + 110592 x^4 + 2016 x^5 - 171360 x^6 + 144 x^7 + 64521 x^8 + x^9$

I cannot find a coeficciant of $\displaystyle x$ because, after expanding the 2 brackets. Upon multiplying them both together, the 2 $\displaystyle x$'s that are in each bracket make an $\displaystyle x^2$ therefore eradicating a coefficiant of $\displaystyle x$

Any help would be greatly appreciated as if this comes up in the exam, I would be prety stuck

thanks!
• Feb 28th 2010, 08:08 AM
CaptainBlack
Quote:

Originally Posted by fishkeeper
Ive done this (I think) and got:

$\displaystyle 512 - 18432 x^2 + 110592 x^4 + 2016 x^5 - 171360 x^6 + 144 x^7 + 64521 x^8 + x^9$

I cannot find a coeficciant of $\displaystyle x$ because, after expanding the 2 brackets. Upon multiplying them both together, the 2 $\displaystyle x$'s that are in each bracket make an $\displaystyle x^2$ therefore eradicating a coefficiant of $\displaystyle x$

Any help would be greatly appreciated as if this comes up in the exam, I would be prety stuck

thanks!

You should know that you don't have to do the full expansions for this type of question?

$\displaystyle (1-2x)^4(2+x)^9=[1+4(-2x) + ... ]2^9[1+9(x/2) +...]=2^9[1-8x+(9/2)x+...]$

where the $\displaystyle ...$ denote terms of higher order than $\displaystyle 1$

So the coefficient you are looking for is: $\displaystyle -2^9(7/2)=-7 \times 2^8$

CB
• Feb 28th 2010, 08:13 AM
Quote:

Originally Posted by fishkeeper
Ive done this (I think) and got:

$\displaystyle 512 - 18432 x^2 + 110592 x^4 + 2016 x^5 - 171360 x^6 + 144 x^7 + 64521 x^8 + x^9$

I cannot find a coeficciant of $\displaystyle x$ because, after expanding the 2 brackets. Upon multiplying them both together, the 2 $\displaystyle x$'s that are in each bracket make an $\displaystyle x^2$ therefore eradicating a coefficiant of $\displaystyle x$

Any help would be greatly appreciated as if this comes up in the exam, I would be prety stuck

thanks!

Hi fishkeeper,

$\displaystyle (1-2x)^4=1+\binom{4}{1}(-2x)+\binom{4}{2}(-2x)^2+....last\ 2\ terms$

$\displaystyle (2+x)^9=2^9+\binom{9}{1}2^8x+\binom{9}{2}2^7x^2+.. .remaining\ terms$

The coefficient of x is found from the constant in one expanded factor multiplied by the coefficient of x in the other.

This is $\displaystyle (1)\binom{9}{1}2^8\ added\ to\ \binom{4}{1}(-2)2^9$
• Mar 1st 2010, 02:22 AM
fishkeeper
Quote:

Originally Posted by CaptainBlack
You should know that you don't have to do the full expansions for this type of question?

$\displaystyle (1-2x)^4(2+x)^9=[1+4(-2x) + ... ]2^9[1+9(x/2) +...]=2^9[1-8x+(9/2)x+...]$

where the $\displaystyle ...$ denote terms of higher order than $\displaystyle 1$

So the coefficient you are looking for is: $\displaystyle -2^9(7/2)=-7 \times 2^8$

CB

Sorry, but I really don't understand this at all.

Why does it not have to be expanded? And, are you just putting the 2 brackets into the general formula for an arithmetic progression? As it looks vaguely like that?
• Mar 1st 2010, 02:42 AM
Hi fishkeeper,

Higher powers of x will not be involved in computing the $\displaystyle x^1$ term, since multiplications of those will only contribute to higher powers of x.

For instance

$\displaystyle (3+x+x^2)(2)=6+2x+2x^2$

The x term results from the multiplication of x by the constant.

Or..

$\displaystyle (3+x+x^2)(2+x+x^2)=6+3x+3x^2+2x+x^2+x^3+2x^2+x^3+x ^4$

$\displaystyle =6+(3x+2x)+(3x^2+x^2)+(x^3+x^3)+x^4$

The resultant x term is the sum of the constant from one factor multiplied by the x term in the other factor from both sets of factors.

When CaptainBlack wrote out the pair of square brackets multiplied together,
he only wrote out the first few terms of the binomial expansions, as higher powers of x will not contribute to the $\displaystyle x^1$ term.
That's all the question wanted you to discover.
The question wanted you to see that only a pair of terms from both expansions contribute to the resultant x term.

CaptainBlack placed the first few terms of the binomial expansions of both $\displaystyle (1-2x)^4$ and $\displaystyle (2+x)^9$ into the square brackets, as it is unnecessary to write the rest of them to obtain the coefficient of the $\displaystyle x^1$ term only.
if you needed the entire expansion, then all the terms of both would be required.
• Mar 1st 2010, 03:20 AM
CaptainBlack
Quote:

Originally Posted by Archie Meade
Hi fishkeeper,

Higher powers of x will not be involved in computing the $\displaystyle x^1$ term, since multiplications of those will only contribute to higher powers of x.

For instance

$\displaystyle (3+x+x^2)(2)=6+2x+2x^2$

The x term results from the multiplication of x by the constant.

Or..

$\displaystyle (3+x+x^2)(2+x+x^2)=6+3x+3x^2+2x+x^2+x^3+2x^2+x^3+x ^4$

$\displaystyle =6+(3x+2x)+(3x^2+x^2)+(x^3+x^3)+x^4$

The resultant x term is the sum of the constant from one factor multiplied by the x term in the other factor from both sets of factors.

When CaptainBlack wrote out the pair of square brackets multiplied together,
he only wrote out the first few terms of the binomial expansions, as higher powers of x will not contribute to the $\displaystyle x^1$ term.
That's all the question wanted you to discover.
The question wanted you to see that only a pair of terms from both expansions contribute to the resultant x term.

CaptainBlack placed the first few terms of the binomial expansions of both $\displaystyle (1-2x)^4$ and $\displaystyle (2+x)^9$ into the square brackets, as it is unnecessary to write the rest of them to obtain the coefficient of the $\displaystyle x^1$ term only.
if you needed the entire expansion, then all the terms of both would be required.

Another point is that a full expansion could be impractical but the question could still be answered by the method given:

For example find the coeficient of x in the expansion of:

$\displaystyle (1+4x)^{1025} (3-2x)^{77}$

CB