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Help with Binomial Expansion question.

Hi, I did this a few weeks ago, but now I have come back to it I feel I am doing the right thing but I am getting the wrong answer everytime I try it.

Expand (1 + x + x²)^3 as far as the x^3 term.

I did this: [ 1 + (x + x²)]^3

= 1 + [x^3] + [3x²(x²)] + [3x(x²)²] + (x²)^3

And when I simplify this (excluding the terms with powers greater than 3), I get the wrong answer. Have I done something wrong or is the answer wrong? I would give you the answer so you can tell me if its right, but my book isn't currently with me, sorry.
Reply 1
Avatar for raheem94
raheem94
13
Original post by ERdoctor
Hi, I did this a few weeks ago, but now I have come back to it I feel I am doing the right thing but I am getting the wrong answer everytime I try it.

Expand (1 + x + x²)^3 as far as the x^3 term.

I did this: [ 1 + (x + x²)]^3

= 1 + [x^3] + [3x²(x²)] + [3x(x²)²] + (x²)^3

And when I simplify this (excluding the terms with powers greater than 3), I get the wrong answer. Have I done something wrong or is the answer wrong? I would give you the answer so you can tell me if its right, but my book isn't currently with me, sorry.


Where did you use binomial expansion here?
Reply 2
Avatar for ERdoctor
ERdoctor
OP
9
SORRY!

I think I know what i did wrong. I should have factorised the (x + x²)^3 bracket to get a 1 for the first term, other wise the expansion won't work. My mistake!

Am I right?
Have you multiplied the terms by the binomial coefficients as well? You know, the ones you get off Pascal's triangle?
Reply 4
Avatar for ERdoctor
ERdoctor
OP
9
Original post by raheem94
Where did you use binomial expansion here?


I didn't show my full working out, but I did use binomial expansion. I think I know where I got it wrong though, I should have factorised more and got a 1 as my first term instead of X.

So rather than having [1 + (x + x²)]^3

I would have [1 + x(1 + x)]^3 Before expanding binomially.

Is that right?
Reply 5
Avatar for raheem94
raheem94
13
Original post by ERdoctor
SORRY!

I think I know what i did wrong. I should have factorised the (x + x²)^3 bracket to get a 1 for the first term, other wise the expansion won't work. My mistake!

Am I right?


Just use the formula,
1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+..... 1 +nx + \dfrac{n(n-1)}{2!}x^2 + \dfrac{n(n-1)(n-2)}{3!}x^3 +.....

Where n=3 and x=(x+x^2)
Ahhh, so you do two separate binomial expansions, and then add their x^3 terms together... I think that could work.
Reply 7
Avatar for raheem94
raheem94
13
Original post by ERdoctor
I didn't show my full working out, but I did use binomial expansion. I think I know where I got it wrong though, I should have factorised more and got a 1 as my first term instead of X.

So rather than having [1 + (x + x²)]^3

I would have [1 + x(1 + x)]^3 Before expanding binomially.

Is that right?


You should use, [1 + (x + x²)]^3, i have tried it and i get the right answer.

Spoiler

(edited 12 years ago)
Reply 8
Avatar for ERdoctor
ERdoctor
OP
9
Original post by raheem94
Just use the formula,
1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+..... 1 +nx + \dfrac{n(n-1)}{2!}x^2 + \dfrac{n(n-1)(n-2)}{3!}x^3 +.....

Where n=3 and x=(x+x^2)


Yeah, that worked. I knew I was over complicating it again.

Thanks! My answer was 1 + 3x +6x² + 7x^3
Reply 9
Avatar for raheem94
raheem94
13
Original post by ERdoctor
Yeah, that worked. I knew I was over complicating it again.

Thanks! My answer was 1 + 3x +6x² + 7x^3


Your answer is correct :smile:.

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