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 :)