Expand Polynomials: Finding Coefficients Using Pascal's Triangle

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Homework Statement

Expand (a+b)ⁿ

Homework Equations

The Attempt at a Solution

Substituting n=2, (a+b)ⁿ = a² + 2ab + b²
Substituting n=3, (a+b)ⁿ = a³ + 3a²b + 3ab² + b³

It's easy to see the powers of a decrease at the same time as the powers of b increase by order 1 each time. Substituting n-1 and n-2 into my attempt at a general form yields

aⁿ + k₁a^n-1b^n-2 + ... + k₂a^n-2b^n-1 + bⁿ

(where k is the co-efficient of a term in the series)

I can use pascals triangle to find the co-efficients quite easily, however I'm struggling to find a pattern that I can write mathematically where I can substitute some value for n and output some value for k.

 

[Mentallic]

We use the term combinatorial denoted by $$^nC_k$$ or $$C(n,k)$$ where n is the power in the expansion and k is the k^th coefficient in the expansion, starting from k=0 for the aⁿb⁰ factor and ending in k=n for the a⁰bⁿ factor.

You can find the value of each combinatorial by either looking at the n^th row in pascals triangle and the k^th number across (again, the 1 at the start being k=0) or you can use a calculator...
or... since you're looking for a more mathematical approach to solving each combinatorial:

$$^nC_k=\frac{n!}{k!(n-k)!}$$

 

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We use the term combinatorial denoted by $$^nC_k$$ or $$C(n,k)$$ where n is the power in the expansion and k is the k^th coefficient in the expansion, starting from k=0 for the aⁿb⁰ factor and ending in k=n for the a⁰bⁿ factor.

You can find the value of each combinatorial by either looking at the n^th row in pascals triangle and the k^th number across (again, the 1 at the start being k=0) or you can use a calculator...
or... since you're looking for a more mathematical approach to solving each combinatorial:

$$^nC_k=\frac{n!}{k!(n-k)!}$$

Very good. Thanks.

What's the norm in writing this as an infinite sum? Do we write

Sigma k=0 to n ⁿC_k a^n-1 b^n-2

 

[Mentallic]

As an infinite sum? I think you mean just a sum :tongue:

$$\sum_{k=0}^n^nC_ka^{n-k}b^k$$

 

[Mentallic]

Oh by the way, I didn't notice it until you wrote out your sum like that in post #3. You expanded the general case incorrectly -

it's meant to be

$$a^n+^nC_1a^{n-1}b+^nC_2a^{n-2}b^2+...+^nC_{n-2}a^2b^{n-2}+^nC_{n-1}ab^{n-1}+b^n$$

Note that $$^nC_0=^nC_n=1$$ and ofcourse $$a^0=b^0=1$$ just so you know why the first and last term are simply aⁿ and bⁿ respectively.

 

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As an infinite sum? I think you mean just a sum :tongue:

$$\sum_{k=0}^n^nC_ka^{n-k}b^k$$

Haha, yep. I was thinking to infinity, nicely picked up

Oh by the way, I didn't notice it until you wrote out your sum like that in post #3. You expanded the general case incorrectly -

it's meant to be

$$a^n+^nC_1a^{n-1}b+^nC_2a^{n-2}b^2+...+^nC_{n-2}a^2b^{n-2}+^nC_{n-1}ab^{n-1}+b^n$$

Note that $$^nC_0=^nC_n=1$$ and ofcourse $$a^0=b^0=1$$ just so you know why the first and last term are simply aⁿ and bⁿ respectively.

Aah, the $$a^{n-k}$$ and $$b^{k}$$ makes sense :D [STRIKE]and the first and last terms made sense already :D[/STRIKE] well, the way I thought about it was just by simply looking at the pattern when I substituted n=2 and n=3. I naturally saw that n will remain on the first and last terms. It got me through (in a kind of a hack) but even better to know the right way about finding the first and last terms in the series. Sometimes leaving off the invisible 1 is misleading.

Thanks heaps for your help.

 

[Mentallic]

No worries