# Binomial Expansion

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• August 6th 2010, 05:56 AM
Punch
Binomial Expansion
Expand the first 3 terms in $[6-13x+6x^2]^5$
• August 6th 2010, 06:38 AM
yeKciM
Quote:

Originally Posted by Punch
Expand the first 3 terms in $[6-13x+6x^2]^5$

when using expansion :D of $(a+b)^n$

$T_{k+1}=\displaystyle \binom {n}{k} a^{n-k} \cdot b^{k}$

that's to find any member :D (first, second, ... )

so u try :D and see where do u go with this :D
• August 6th 2010, 06:39 AM
Unknown008
What you can do is treat it like this:

$[6 - (13x - 6x^2)]^5$

Then apply the formula to find the first three terms.

$(a + b)^n = a^n + \binom{n}{1} (a)^{n-1}(b) + \binom{n}{2} (a)^{n-2}(b)^2+...$

Taking a = 6
b = - (13x - 6x^2)
• August 6th 2010, 06:45 AM
Punch
that makes sense thanks
• August 6th 2010, 06:46 AM
yeKciM
Quote:

Originally Posted by Unknown008
What you can do is treat it like this:

$[6 - (13x - 6x^2)]^5$

Then apply the formula to find the first three terms.

$(a + b)^n = a^n + \binom{n}{1} (a)^{n-1}(b) + \binom{n}{2} (a)^{n-2}(b)^2+...$

Taking a = 6
b = - (13x - 6x^2)

perhaps that's the better way to go :D

formula that i posted is the same, but way that it's wrote it's way easier to find let's say 181 element of some expansion :D :D:D:D not for this problem but works the same :D
• August 6th 2010, 06:56 AM
Plato
Quote:

Originally Posted by Punch
Expand the first 3 terms in $[6-13x+6x^2]^5$

I have a question about the question itself.
Who is to say which are the first three terms?
• August 6th 2010, 07:02 AM
Unknown008
Quote:

Originally Posted by Plato
I have a question about the question itself.
Who is to say which are the first three terms?

Yes, the question should have included something like: "the first three terms in ascending powers of x"
• August 6th 2010, 07:03 AM
undefined
Quote:

Originally Posted by Plato
I have a question about the question itself.
Who is to say which are the first three terms?

I was wondering that too, figured the only natural choices were: (1) constant term, x^1 term, x^2 term; (2) x^10 term, x^9 term, x^8 term; and that since the trinomial we are expanding is given in form (1) it would be acceptable to assume form (1).
• August 6th 2010, 07:12 AM
Quote:

Originally Posted by Punch
Expand the first 3 terms in $[6-13x+6x^2]^5$

You can also expand

$\left(x[6x-13]+6\right)^5$

It may be more compact.
• August 6th 2010, 07:16 AM
Soroban
Hello, Punch!

This is a messy one . . . even with a good procedure.

Quote:

$\text{Find the first 3 terms of: }\;(6-13x+6x^2)^5$

It is evident that the first three terms are:
. . . . $\text{constant},\;x\text{-term},\;x^2\text{-term}$

$\text{We have: }\:(a + b + c)^5\quad \text{where: }\:\begin{Bmatrix}a &=& 6 \\ b &=& \text{-}13x \\ c &=& 6x^2 \end{Bmatrix}$

The "Trinomial Expansion" forms an array:

$\begin{array}{cccccccccccc}&&& {5\choose5,0,0}a^5 \\ && {5\choose4,1,0}a^4b && {5\choose4,0,1}a^4c \\ & {5\choose3,2,0}a^3b^2 && {5\choose3,1,1}a^3bc && {5\choose3,0,2}a^3c^2 \\ {5\choose2,3,0}a^2b^3 && {5\choose2,2,1}a^2b^2c && {5\choose2,1,2}a^2bc^2 && {5\choose2,0,3}a^2c^3 \\ \vdots && \vdots && \vdots && \vdots\end{array}$

We want only the first four terms (reading from the top).
. . [The rest contain $x^3$ or higher powers.]

. . ${5\choose5,0,0}(6^5) + {5\choose4,1,0}(6^4)(\text{-}13x) + {5\choose4,0,1}(6^4)(6x^2) + {5\choose3,2,0}(6^3)(\text{-}13x)^2$

. . $=\;1(7776) + 5(1296)(\text{-}13x) + 5(1296)(6x^2) + 10(216)(169x^2)$

. . $=\;7,\!776 - 84,\!240x + 38,\!880x^2 + 365,\!040x^2$

. . $=\;7,\!776 - 84,\!240x + 403,\!920x^2$

But check my work . . . please!
• August 6th 2010, 07:20 AM
yeKciM
or look at it like this

$(6-13x+6x^2)^5=[-x+(x-1)(6x-6)]^5$

and so when doing u'll get

$T_1= \displastyle \binom {5}{0} (-x)^5 [(x-1)(6x-6)]^0=-x^5$

$T_2= \displastyle \binom {5}{1} (-x)^4 [(x-1)(6x-6)]^1=5x^4[6x^2-12x+6]=30x^6-60x^5+30x^4$

$T_3= \displastyle \binom {5}{2} (-x)^3 [(x-1)(6x-6)]^2=-10x^3[(x-1)^2(6x-6)^2]=-360x^7+1440x^6-2160x^5+1440x^4-360x^3$

u can go further... but u get the idea :D

or am i wrong here :D ?
• August 6th 2010, 07:38 AM
undefined
Quote:

Originally Posted by Soroban
. . $=\;7,\!776 - 84,\!240x + 403,\!920x^2$

But check my work . . . please!

Confirmed with Mathematica. Unknown008's method gives the same result.
• August 6th 2010, 07:42 AM
undefined
Quote:

Originally Posted by yeKciM
or look at it like this

$(6-13x+6x^2)^5=[-x+(x-1)(6x-6)]^5$

and so when doing u'll get

$T_1= \displastyle \binom {5}{0} (-x)^5 [(x-1)(6x-6)]^0=-x^5$

$T_2= \displastyle \binom {5}{1} (-x)^4 [(x-1)(6x-6)]^1=5x^4[6x^2-12x+6]=30x^6-60x^5+30x^4$

$T_3= \displastyle \binom {5}{2} (-x)^3 [(x-1)(6x-6)]^2=-10x^3[(x-1)^2(6x-6)^2]$
$=-360x^7+1440x^6-2160x^5+1440x^4-360x^3$

u can go further... but u get the idea :D

or am i wrong here :D ?

This makes it harder to find the first three terms, compared with what Unknown008 posted, doesn't it?
• August 6th 2010, 07:50 AM
yeKciM
Quote:

Originally Posted by undefined
This makes it harder to find the first three terms, compared with what Unknown008 posted, doesn't it?

hmmmm :D

probably lost in translation, but if u look at $(6-13x+6x^2)^5=[-x+(x-1)(6x-6)]^5$ then first one should be :

$T_1=\displaystyle \binom{5}{0} a^{5-0}\cdot b^0$

and so on...

but probably i'm little (or a lot) of with meaning of "terms" for me it's like n-th member or something like that :D

Edit: Sorry, but now i'm little a bit of confused :D
• August 6th 2010, 07:57 AM
undefined
Quote:

Originally Posted by yeKciM
hmmmm :D

probably lost in translation, but if u look at $(6-13x+6x^2)^5=[-x+(x-1)(6x-6)]^5$ then first one should be :

$T_1=\displaystyle \binom{5}{0} a^{5-0}\cdot b^0$

and so on...

but probably i'm little (or a lot) of with meaning of "terms" for me it's like n-th member or something like that :D

Edit: Sorry, but now i'm little a bit of confused :D

There are 11 terms in the expansion of $(6-13x+6x^2)^5$, starting with the constant term and ending with the $x^{10}$ term. Soroban gave the first three terms in post #10.
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