# Laws of Binomial theorem (2.1)

Greetings everyone, welcome to the part 2.1 of the Laws of Binomial theorem(2.2 coming next week). In this discussion we will use integration and differentiation to transform our original equation to get a desired series . Check the 1st part before you continue with this one. We are starting with the binomial expansion of $(1+x)^{n}$ . We are calling it equation $I$ . Where $x$ is a complex number and $n$ is a whole number.

$\ (1+x)^{n} = \dbinom{n}{0}x^{0}+\dbinom{n}{1}x^{1}+\dbinom{n}{2}x^{2}+\cdots+\dbinom{n}{n-1}x^{n-1}+\dbinom{n}{n}x^{n} = \displaystyle\sum_{r=0}^{n}\dbinom{n}{r}x^{r}$

Some Important results obtained from $I$ are

$1)$ put $x=1$

$\boxed{ 2^{n} = \dbinom{n}{0}+\dbinom{n}{1}+\dbinom{n}{2}+\cdots+\dbinom{n}{n-1}+\dbinom{n}{n} = \displaystyle\sum_{r=0}^{n}\dbinom{n}{r}2^{r} }$

$2)$ put $x=-1$

$0 = \dbinom{n}{0}-\dbinom{n}{1}+\dbinom{n}{2}-\cdots+\dbinom{n}{n-1}(-1)^{n-1}+\dbinom{n}{n}(-1)^{n} = \displaystyle\sum_{r=0}^{n}\dbinom{n}{r}(-1)^{r}$

$\boxed{ \dbinom{n}{0}+\dbinom{n}{2}+\dbinom{n}{4}+\cdots=\dbinom{n}{1}+\dbinom{n}{3}+\dbinom{n}{5}+\cdots = 2^{n-1} }$

Now , Differentiate $I$ with respect to $x$

$n(1+x)^{n-1} =1\cdot\dbinom{n}{1}+2\cdot\dbinom{n}{2}x^{1}+\cdots+(n-1)\cdot\dbinom{n}{n-1}x^{n-2}+n\cdot\dbinom{n}{n}x^{n-1} = \displaystyle\sum_{r=0}^{n}r\dbinom{n}{r}x^{r-1}$

put $x=1$

$\boxed{ n\cdot2^{n-1} =1\cdot\dbinom{n}{1}+2\cdot\dbinom{n}{2}+\cdots+(n-1)\cdot\dbinom{n}{n-1}+n\cdot\dbinom{n}{n} = \displaystyle\sum_{r=0}^{n}r\dbinom{n}{r} }$

Similarly we can differentiate $I$ twice to get a new series. Let's Try to find the sum of

$1^2\cdot\dbinom{n}{1}+2^2\cdot\dbinom{n}{2}+\cdots+(n-1)^2\cdot\dbinom{n}{n-1}+n^2\cdot\dbinom{n}{n} = \displaystyle\sum_{r=0}^{n}r^2\dbinom{n}{r}$

Let's start with $I$ $\ (1+x)^{n} = \dbinom{n}{0}x^{0}+\dbinom{n}{1}x^{1}+\cdots+\dbinom{n}{n}x^{n} = \displaystyle\sum_{r=0}^{n}\dbinom{n}{r}x^{r} \quad\quad\quad\quad\quad\quad\quad\quad \small\color{#3D99F6}{\text{0th derivative}}$

$n(1+x)^{n-1} =1\cdot\dbinom{n}{1}+2\cdot\dbinom{n}{2}x^{1}+\cdots+n\cdot\dbinom{n}{n}x^{n-1} = \displaystyle\sum_{r=0}^{n}r\dbinom{n}{r}x^{r-1} \quad \small\color{#3D99F6}{\text{1st derivative}}$

$nx(1+x)^{n-1} =1\cdot\dbinom{n}{1}x^{1}+2\cdot\dbinom{n}{2}x^{2}+\cdots+n\cdot\dbinom{n}{n}x^{n} = \displaystyle\sum_{r=0}^{n}r\dbinom{n}{r}x^{r} \quad \small\color{#3D99F6}{\text{times x on both sides}}$

$\underbrace{n(1+x)^{n-1}+n(n-1)x(1+x)^{n-2}}_{n(nx+1)(1+x)^{n-2}} =1^2\cdot\dbinom{n}{1}+2^2\cdot\dbinom{n}{2}x^1+\cdots+n^2\cdot\dbinom{n}{n}x^{n-1} = \displaystyle\sum_{r=0}^{n}r^2\dbinom{n}{r}x^{r-1} \quad \small\color{#3D99F6}{\text{2nd derivative}}$

$\boxed{ n(n+1)2^{n-2}=1^2\cdot\dbinom{n}{1}+2^2\cdot\dbinom{n}{2}+\cdots+n^2\cdot\dbinom{n}{n}= \displaystyle\sum_{r=0}^{n}r^2\dbinom{n}{r} \quad \small\color{#3D99F6}{\text{Put x=1}} }$

For Practice try to find the sum of the following series:

$1) \quad 1^3\cdot\dbinom{n}{1}+2^3\cdot\dbinom{n}{2}+\cdots+(n-1)^3\cdot\dbinom{n}{n-1}+n^3\cdot\dbinom{n}{n}$

$2) \quad 1\cdot2\cdot\dbinom{n}{2}+3\cdot2\cdot\dbinom{n}{3}+\cdots+(n-1)\cdot(n-2)\cdot\dbinom{n}{n-1}+n\cdot(n-1)\dbinom{n}{n}$

Thank You :)

Note by Sabhrant .
2 years, 9 months ago

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Woah, these identities are really nice and useful! I can see combinatorial proofs being used here.

I think you should create a wiki page to showcase these few identities and how to extrapolate these pattern.

- 2 years, 8 months ago

@Sambhrant Sachan Agreed! Let me know if you need help getting started. Several of these could be added directly to binomial theorem.

Staff - 2 years, 8 months ago

As soon as i finish my series , i will work on the wiki :) .

- 2 years, 8 months ago

Sir, can you confirm that total partitions (with permutations) of a integer $N$ is $2^{N-1}$ .

For example ,

\begin{aligned} 3 & = 1+1+1 \quad \quad 2^{3-1} = \boxed{4}\\ & = 1+2 \\ & = 2+1 \\ & = 3 \\ 4 & = 1+1+1+1 \quad \quad 2^{4-1} = \boxed{8} \\ & = 1+1+2 \\ & = 1+2+1 \\ & = 2+1+1 \\ & = 1+3 \\ & = 3+1 \\ & = 2+2 \\ & = 4 \end{aligned}

If it is correct i will explain the proof in combinatorics .

- 2 years, 8 months ago

Yes it is. It follows directly from product rule.

I think the product rule wiki or the bijection, injection, and surjection wiki would be a much better place for it.

Staff - 2 years, 8 months ago