• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

Binomial theorem question (1 Viewer)

fashionista

Tastes like chicken
Joined
Nov 29, 2003
Messages
900
Location
iN ur PaNTs
Gender
Undisclosed
HSC
N/A
arrgh this one looks so simple and small but the only way i know how to do it is taking soooo long
please help me!!
by means of factors of (x^2-3x+2) expand (x^2-3x+2)^6 in ascending powers of x as far as the term x^3

so i did(x-2)^6(x-1)^6 and then im getting all the terms together and its taking so long, theres gotta be an easier way.
please help!!
 

mojako

Active Member
Joined
Mar 27, 2004
Messages
1,333
Gender
Male
HSC
2004
fashionista said:
arrgh this one looks so simple and small but the only way i know how to do it is taking soooo long
please help me!!
by means of factors of (x^2-3x+2) expand (x^2-3x+2)^6 in ascending powers of x as far as the term x^3

so i did(x-2)^6(x-1)^6 and then im getting all the terms together and its taking so long, theres gotta be an easier way.
please help!!
The approach I can think of atm:
Find the general Tn formula for (x-2)^6 and for (x-1)^6.
Then find the term with x^0, with x^1, x^2, x^3.
Example with x^3:
Code:
       (x-2)<sup>6</sup>(x-1)<sup>6</sup>
power:   0     3 (so u want x^0 from the (x-2)^6, and x^3 from the other)
         1     2
         2     3
         3     0
 
Last edited:

sub

Member
Joined
Apr 19, 2004
Messages
621
hey mojako, dont u also have to list out x^2, x, and constants? cos it says as far the power x^3...?
 

mojako

Active Member
Joined
Mar 27, 2004
Messages
1,333
Gender
Male
HSC
2004
sub said:
hey mojako, dont u also have to list out x^2, x, and constants? cos it says as far the power x^3...?
yea.. I said x^3 as "example" in the edited post.
can you think of a better approach? mine still takes significant work.
 

fashionista

Tastes like chicken
Joined
Nov 29, 2003
Messages
900
Location
iN ur PaNTs
Gender
Undisclosed
HSC
N/A
thank u mojako! i was doing it your way too, i think that's the only way you can do it.

i have another question tho :shy:
by expanding both sides of the identity (1+x)^(m+3)=(1+x)^m(1+x)^3 prove that
(m+n)Cr=mCr +3{mC(r-1)]+ 3[mC(r-2)] + mC(r-3)
i expanded the RHS of the identitybut im confused about getting the bit you have to prove, i just dont get why it turns out the way it's supposed to be proven.

no no no forget it..it's ok i get it now!!
 
Last edited:

mojako

Active Member
Joined
Mar 27, 2004
Messages
1,333
Gender
Male
HSC
2004
well firstly its (m+3)Cr not (m+n)Cr
now:
(1+x)<sup>m+3</sup>=(1+x)<sup>m</sup>(1+x)<sup>3</sup>

general term of LHS
= (m+3)Cr x<sup>r</sup>

general term of RHS
= mCk x<sup>k</sup> (1+3x+3x<sup>2</sup>+x<sup>3</sup>)
= mCk x<sup>k</sup> + 3x mCk x<sup>k</sup> + 3x<sup>2</sup> mCk x<sup>k</sup> + x<sup>3</sup> mCk x<sup>k</sup>

now you want the term with x<sup>r</sup> on RHS because then you can equate the coefficient with the LHS one

and this gives you the result.

grrr.. wasted my time... :)
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top