Waste less time on Facebook — follow Brilliant.
×

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 \( I \) \[ \ (1+x)^{n} = \dbinom{n}{0}x^{0}+\dbinom{n}{1}x^{1}+\cdots+\dbinom{n}{n}x^{n} = \displaystyle\sum_{r=0}^{n}\dbinom{n}{r}x^{r} \quad\quad\quad\quad\quad\quad\quad\quad \small\color{blue}{\text{0th derivative}} \]

\[ n(1+x)^{n-1} =1\cdot\dbinom{n}{1}+2\cdot\dbinom{n}{2}x^{1}+\cdots+n\cdot\dbinom{n}{n}x^{n-1} = \displaystyle\sum_{r=0}^{n}r\dbinom{n}{r}x^{r-1} \quad \small\color{blue}{\text{1st derivative}} \]

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

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

\[\boxed{ n(n+1)2^{n-2}=1^2\cdot\dbinom{n}{1}+2^2\cdot\dbinom{n}{2}+\cdots+n^2\cdot\dbinom{n}{n}= \displaystyle\sum_{r=0}^{n}r^2\dbinom{n}{r} \quad \small\color{blue}{\text{Put x=1}} } \]

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 Sachan
11 months, 2 weeks ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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 - 11 months, 1 week ago

Log in to reply

@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 - 11 months ago

Log in to reply

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

For example ,

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

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

Sabhrant Sachan - 11 months ago

Log in to reply

@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 - 11 months ago

Log in to reply

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

Sabhrant Sachan - 11 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...