What is the proof of the Binomial Theorem, other than the induction method? How can we find the expansion of binomails with indices like 2n, 3n, 4n..?

Question

1. The proof of the binomial theorem is almost always given with induction. Is there any other way to proof the binomial theorem. I have been trying to find it, but have not yet done it.
2. What is the expansion of $((1+x)^n)^2$. Is there any way of finding the expansions for similar formulas like $((1+x)^n)^3$, $((1+x)^4)^n$,........., $((1+x)^n)^n).$

Will simplifying the terms like $(1+x)^n*(1+x)^n$ and then multiplying the general terms of both the expansions help?

Answer 1

For your first question we can also show it using the Taylor series formula

$$f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}x^k\ .$$

Fix $n\in\mathbb{N}$ and let $f(x) = (1+x)^n$. Then $f$ is analytic (it is just a polynomial) and so we can apply the above formula. We only need to compute the $k$th derivative at $0$. For $k\leq n$

$$f^{(k)}(x)= n\times(n-1)\times(n-2)\times\cdots\times(n-k+1)\times (1+x)^{n-k} =\frac{n!}{(n-k)!}(1+x)^{n-k}\ ,$$

while for $k> n$ we have $$f^{(k)}(x)=0\ .$$ Maybe you can say this step needs induction, but it's certainly more clear than the whole formula. Plugging in $x=0$ we see

$$f^{(k)}(0)=\begin{cases}\frac{n!}{(n-k)!}& k\leq n\\ 0 & k > n\end{cases}$$

Inserting this back into the Taylor series formula gives

$$f(x) = \sum_{k=0}^n \frac{n!}{(n-k)!k!}x^k = \sum_{k=0}^n \begin{pmatrix}n\\k\end{pmatrix}x^k$$

Edit: To answer your second question $((1+x)^n)^m = (1+x)^{nm}$ and so you can just replace all the $n$'s by $nm$'s in the binomial theorem to get the answer.

Answer 2

Here is generating function approach. We show the validity of \begin{align*} (1+x)^n=\sum_{k=0}^n\binom{n}{k}x^k\tag{1} \end{align*} by showing that left-hand side and right-hand side have the same exponential generating function. \begin{align*} \color{blue}{e^{(1+x)z}} \end{align*}

We start with the exponential generating function of the left-hand side. We obtain \begin{align*} \sum_{n=0}^\infty \color{blue}{(1+x)^n}\frac{z^n}{n!}&=\sum_{n=0}^\infty\frac{\left((1+x)z\right)^n}{n!}\\ &=e^{(1+x)z}\tag{2.1}\\ &=e^{xz}e^z\tag{2.2}\\ &=\left(\sum_{k=0}^\infty \frac{(xz)^k}{k!}\right)\left(\sum_{l=0}^\infty\frac{z^l}{l!}\right)\tag{2.3}\\ &=\sum_{n=0}^\infty\left(\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{x^k}{k!}\,\frac{1}{l!}\right)z^n\tag{2.4}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac{x^k}{k!}\,\frac{1}{(n-k)!}\right)z^n\tag{2.5}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac{n!}{k!}\,\frac{1}{(n-k)!}x^k\right)\frac{z^n}{n!}\tag{2.6}\\ &\,\,=\sum_{n=0}^\infty\left(\color{blue}{\sum_{k=0}^n\binom{n}{k}x^k}\right)\frac{z^n}{n!}\tag{2.7}\\ \end{align*} and the claim (1) follows.

Comment:

• In (2.1) we use the definition of the exponential series.

• In (2.2) we apply the rule $e^{p+q}=e^pe^q$.

• In (2.3) we use again the definition as in (2.1).

• In (2.4) we rearrange according increasing powers of $z$.

• In (2.5) we eliminate $l$ by substituting $l\to n-k$.

• In (2.6) we expand numerator and denominator with $n!$.

• In (2.7) we use $\binom{n}{k}=\frac{n!}{k!(n-k)!}$.