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 (1+x)^{n} . We are calling it equation I I . Where xx is a complex number and n n is a whole number.

 (1+x)n=(n0)x0+(n1)x1+(n2)x2++(nn1)xn1+(nn)xn=r=0n(nr)xr \ (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 I are

1)1) put x=1 x=1

2n=(n0)+(n1)+(n2)++(nn1)+(nn)=r=0n(nr)2r \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)2) put x=1 x=-1

0=(n0)(n1)+(n2)+(nn1)(1)n1+(nn)(1)n=r=0n(nr)(1)r 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}

(n0)+(n2)+(n4)+=(n1)+(n3)+(n5)+=2n1 \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 I with respect to x x

n(1+x)n1=1(n1)+2(n2)x1++(n1)(nn1)xn2+n(nn)xn1=r=0nr(nr)xr1 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 x=1

n2n1=1(n1)+2(n2)++(n1)(nn1)+n(nn)=r=0nr(nr) \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 I twice to get a new series. Let's Try to find the sum of

12(n1)+22(n2)++(n1)2(nn1)+n2(nn)=r=0nr2(nr) 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 I  (1+x)n=(n0)x0+(n1)x1++(nn)xn=r=0n(nr)xr0th derivative \ (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)n1=1(n1)+2(n2)x1++n(nn)xn1=r=0nr(nr)xr11st 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)n1=1(n1)x1+2(n2)x2++n(nn)xn=r=0nr(nr)xrtimes x on both sides 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}}

n(1+x)n1+n(n1)x(1+x)n2n(nx+1)(1+x)n2=12(n1)+22(n2)x1++n2(nn)xn1=r=0nr2(nr)xr12nd derivative \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}}

n(n+1)2n2=12(n1)+22(n2)++n2(nn)=r=0nr2(nr)Put x=1\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)13(n1)+23(n2)++(n1)3(nn1)+n3(nn)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)12(n2)+32(n3)++(n1)(n2)(nn1)+n(n1)(nn)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 Sachan
4 years, 6 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.

Pi Han Goh - 4 years, 5 months ago

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

Calvin Lin Staff - 4 years, 5 months ago

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

Sabhrant Sachan - 4 years, 5 months ago

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Sir, can you confirm that total partitions (with permutations) of a integer N N is 2N1 2^{N-1} .

For example ,

3=1+1+1231=4=1+2=2+1=34=1+1+1+1241=8=1+1+2=1+2+1=2+1+1=1+3=3+1=2+2=4 \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 .

Sabhrant Sachan - 4 years, 5 months ago

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@Sabhrant Sachan 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.

Calvin Lin Staff - 4 years, 5 months ago

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