Expand the first 3 terms in $\displaystyle [6-13x+6x^2]^5$

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- Aug 6th 2010, 05:56 AMPunchBinomial Expansion
Expand the first 3 terms in $\displaystyle [6-13x+6x^2]^5$

- Aug 6th 2010, 06:38 AMyeKciM
- Aug 6th 2010, 06:39 AMUnknown008
What you can do is treat it like this:

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

Then apply the formula to find the first three terms.

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

Taking a = 6

b = - (13x - 6x^2) - Aug 6th 2010, 06:45 AMPunch
that makes sense thanks

- Aug 6th 2010, 06:46 AMyeKciM
- Aug 6th 2010, 06:56 AMPlato
- Aug 6th 2010, 07:02 AMUnknown008
- Aug 6th 2010, 07:03 AMundefined
- Aug 6th 2010, 07:12 AMArchie Meade
- Aug 6th 2010, 07:16 AMSoroban
Hello, Punch!

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

Quote:

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

It is evident that the first three terms are:

. . . . $\displaystyle \text{constant},\;x\text{-term},\;x^2\text{-term}$

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

The "Trinomial Expansion" forms an array:

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

We want only the first four terms (reading from the top).

. . [The rest contain $\displaystyle x^3$ or higher powers.]

. . $\displaystyle {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 $

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

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

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

But check my work . . .*please!* - Aug 6th 2010, 07:20 AMyeKciM
or look at it like this

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

and so when doing u'll get

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

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

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

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

or am i wrong here :D ? - Aug 6th 2010, 07:38 AMundefined
- Aug 6th 2010, 07:42 AMundefined
- Aug 6th 2010, 07:50 AMyeKciM

hmmmm :D

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

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

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 - Aug 6th 2010, 07:57 AMundefined