Proving the Binomial Theorem?

[CheddarCheese]

Hello all, first off, I hope I'm posting this question in the right place. If not, I'm sorry, and I'll try not to make the same mistake later.

Well, I've been working on Michael Spivak's Calculus Chapter 2, and am stuck on a question. It asks:

Prove the "binomial theorem": If a and b are any numbers and n is a natural number, then:
(a + b)ⁿ = aⁿ + (nC1)a^n-1b + (nC2)a^n-2b² + ... + (nC(n-1))ab^n-1 + bⁿ = Σ ⁿ_j=0(nCj)a^n-jb^j

I have not yet been taught how to do something like this, yet I am required to complete this for an assignment. I really have no idea where to begin. Any beginning steps and pointers would be very much appreciated.

Thanks!

 

[pka]

Prove the "binomial theorem": If a and b are any numbers and n is a natural number, then:
(a + b)ⁿ = aⁿ + (nC1)a^n-1b + (nC2)a^n-2b² + ... + (nC(n-1))ab^n-1 + bⁿ = Σ ⁿ_j=0(nCj)a^n-jb^j

To expand \(\displaystyle \left( {a + b} \right)^n = \underbrace {(a + b)(a + b) \cdots (a + b)}_{n\text{ terms}}\) we take one term, either \(\displaystyle a\text{ or }b\), from each term of the product.

Thus the term \(\displaystyle \dbinom{n}{k}a^kb^{n-k}\) means we have chosen
\(\displaystyle a\) from \(\displaystyle k\) of those \(\displaystyle n\) terms which means we have taken \(\displaystyle n-k~~b’s\).

Thus you have \(\displaystyle \left( {a + b} \right)^n = \sum\limits_{k = 0}^n {\dbinom{n}{k}a^k b^{n - k} }\).

 

[CheddarCheese]

Hi, Thanks for the reply.

I'm not sure I understand the connection. I see what you mean in your first statement, but I don't know about the second one. How do those two link up to make the final equation?

 

[pka]

Hi, Thanks for the reply.
I'm not sure I understand the connection. I see what you mean in your first statement, but I don't know about the second one. How do those two link up to make the final equation?

That proof is known as a combinatorial proof.
How does one multiply those n factors together?
Well, we select either an a or a b from each factor.
So the proof describes how each term in the final sum is comprised.

If on the other hand, you are looking for a strict algebraic proof you are in need of fairly advanced properties of binomial coefficients.

 

[lookagain]

Prove the "binomial theorem": If a and b are any numbers and n is a natural number, then:
(a + b)ⁿ = aⁿ + (nC1)a^n-1b + (nC2)a^n-2b² + ... + (nC(n-1))ab^n-1 + bⁿ = Σ ⁿ_j=0(nCj)a^n-jb^{j \(\displaystyle * * \)}

I would use Mathematical Induction:


1) Demonstrate the base case is true. (Here, n = 1).


2) Let f(n) = **
Assume f(k) is true.


3) Using the assumption in 2), show that f(k + 1) is true.


4) State a conclusion to the effect "Thus, by the Principle of
Mathematical Induction, I have proven the Binomial Theorem
is true."

 

[pka]

I would use Mathematical Induction:

To do this proof by induction one needs to know properties of binomial coefficients which are much advanced that the problem itself.

 

[lookagain]

To do this proof by induction one needs to know
properties of binomial coefficients which are much advanced that the
problem itself.

One of the steps in the induction method that I would be using centers on being
able to show that


\(\displaystyle C(n, \ m) + C(n, \ m+1) \ = \ C(n + 1, \ m + 1), \ \ where \)

Suppose m and n are appropriate integers, with \(\displaystyle n \ge 1, \ m \ge 0, \ and \ m \le n \ \ **\)


\(\displaystyle \dfrac{n!}{m![(n - m)!]} \ + \ \dfrac{n!}{(m + 1)![n - (m + 1)]!} \ =\)


\(\displaystyle n!\bigg[\dfrac{1}{m![(n - m)!]} \ + \ \dfrac{1}{(m + 1)![n - (m + 1)]!}\bigg] *** \)



\(\displaystyle \text{The LCD is (m + 1)![(n - m)!]. \ \ Multiply the numerators and }\)

\(\displaystyle \text{denominators of each respective fraction}\)


\(\displaystyle \text{by an appropriate respective expression so that the denominators }\)


\(\displaystyle \text{are the same, and then the fractions can be added together.}\)


\(\displaystyle *** \ = \ n!\bigg[\dfrac{(m + 1) + (n - m)}{(m + 1)![(n - m)!]}\bigg]\)


\(\displaystyle = \ n!\bigg[\dfrac{n + 1}{(m + 1)![(n - m)!]}\bigg]\)


\(\displaystyle = \ \dfrac{(n + 1)!}{(m + 1)![(n - m)!]}\)


\(\displaystyle = \ C(n + 1, \ m + 1)\)


\(\displaystyle \boxed{This \ subpart \ is \ proved.}\)





\(\displaystyle ** \ as \ in \ (a + b)^n \ = \ C(n, \ m)a^nb^m, \ beginning \ with \ expanding \ (a + b)^1.\)