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Huh? (Maths question)

Started by ukdeveloper Nov 29, 2005 at 5:11 PM 5 replies 850+ views

Original Post

Nov 29, 2005 at 5:11 PM

Hi, I'm totally lost with my Maths revision. I'm using this textbook which has questions in the front, and the answers (without working) are at the back. I'm working on the binomial expansion, and there's this real bugger which I just can't seem to figure out. It shouldn't be difficult for you, but it was taught to me by the worse of the two professors I get. The question is to expand using the binomial expansion: (2 + 3b)^4 According to the back of the book, the answer is: 16 + 96b + 216b^2 + 216b^3 + 81b^4 I can get the 16 and the 81b^4 fine, but how in hell's bells did they get the rest of it? I'm clueless. There's no point in asking the professor in question - he really is an idiot, and me along with 200 others in the class think that. So hence why I'm asking here. I think once I master how this example works, I should be able to do the rest. Google isn't being too helpful, nor are my textbooks (oddly enough). Thanks in advance.

Coward

Nov 29, 2005 at 5:18 PM

(2 + 3b)^4 = (2 + 3b)(2 + 3b)(2 + 3b)(2 + 3b) = (9b^2 + 12b + 4)(2 + 3b)(2 + 3b)

It seems you think the expressions says 2^4 + (3b)^4, which is not the case..

When factors of sums are multiplied, 3 sums will be the answer:

(a + b)^2 = a^2 + b^2 + 2ab

jfclavette

Nov 29, 2005 at 5:20 PM

I don't really know what the hell binomial expansion is, but isn't it just:

(2+3b)^4 = (2 + 3b)(2 + 3b)(2 + 3b)(2 + 3b)         = (4 + 12b + 9b^2)(4 + 12b + 9b^2)

We can do this because you multiply each term of the first parenthesis with each term of the second and sum them. We can do first*second and third*fourth at the same time because multiplication is associative.

         = 16 + 48b + 36b^2 + 48b + 144b^2 + 108b^3 + 36b^2 + 108b^3 + 81b^4 //M         = 16 + 96b + 216b^2 + 216b^3 + 81b^4

SamLowry

Nov 29, 2005 at 5:20 PM

Binomial Theorem (Mathworld)

kSquared

Nov 29, 2005 at 5:21 PM

Use the binomial expansion theorem like your book says. This is a straightforward application of theory and really shouldn't be terribly difficult. For reference, the theorem is:

(x + y)ⁿ = Sum{i : 0 --> n | C(n, i) * x^{n - i} * yⁱ}

where Sum{} indicates a summation expression and C(n, r) means "n choose r", or n! / ((n - r)! r!).

Conner McCloud

Nov 29, 2005 at 5:26 PM

If in doubt, you can always just do the multiplication by hand. (2 + 3b)^2^2 isn't too hard to calculate.

The Binomial Theorum. The coefficients can be gotten using Pascal's Triangle. In your case, you want the fourth row of the triangle [1 4 6 4 1], x is 2, and a is 3b.

So, your first term is (1)*(2)^0*(3b)^4=81b^4. Your second term is 4*(2)^1*(3b)^3=216b^3. Then 6*(2)^2*(3b)^2=216b^2, 4*(2)^3*(3b)^1=96b, and 1*(2)^4*(3b)^0=16. Add them all together, and you're there.

CM

_goat

Nov 30, 2005 at 6:07 AM

Extrapolating from kSquared:

You have (2 + 3b)^4. The two "parts" inside the brackets are 2, and 3b.

So we start with powers C and D, where C = 4, and D = 0.
We also need Pascal's Triangle, which for this sequence is 1, 4, 6, 4, 1

2^C = 16. (3b)^D = 1.
Pascal's Triangle: 1
1 * (16 * 1) = 16.

Subtract 1 from C, and add 1 to D. (C = 3, D = 1)

2^C = 8. (3b)^D = 3b.
Pascal's Triangle: 4
4 * (8 * 3b) = 96b

Do it again with the subtracting and the adding. (C now equals 2, so does D)

2^C = 4. (3b)^D = 9b^2.
P.T: 6
6 * (4 * 9b^2) = 216b^2

And again. (C = 1, D = 3)

2^C = 2. (3b)^D = 27b^3.
P.T: 4
4 * (2 * 27b^3) = 216b^3

And again. (C = 0, D = 4)

2^C = 1. (3b)^D = 81b^4
P.T: 1
1 * (1 * 81b^4) = 81b^4

Add them all up:

16 + 96b + 216b^2 + 216b^3 + 81b^4

And you're done.

Edit: Psh, I was FAR too slow for Conner. Damn dinner, you got me again.

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