Binomial

#Combinatorics

Note by Dhruv Joshi
4 years, 3 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

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 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai−1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 ∑i=13 \sum_{i=1}^3
\sin \theta sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

need help

Dhruv Joshi - 4 years, 3 months ago

I here want to address Kartik Sharma for his truly marvelous approach and also to give him the credits for this approach. The solution you'll see below is an exact replication of his solution in one of his problems.

Proposition #1:∑m=1n(∑k=1mak)(nm)(−1)m−1=∑m=0n−1(n−1m)am+1(−1)m\displaystyle \text{Proposition \#1:} \large \sum_{m=1}^n {\left(\sum_{k=1}^m{a_k}\right) \binom{n}{m} {(-1)}^{m-1}} = \sum_{m=0}^{n-1}{\binom{n-1}{m} a_{m+1}{(-1)}^m}

Proof:\text{Proof:}

LHS can be written as -

(n1)a1−(n2)(a1+a2)+(n3)a3−⋯+(−1)n−1(nn)(a1+a2+⋯+an)\displaystyle \binom{n}{1}a_1 - \binom{n}{2}(a_1 + a_2) + \binom{n}{3} a_3 - \cdots + {(-1)}^{n-1} \binom{n}{n} (a_1 + a_2 + \cdots + a_n)

=((n1)−(n2)+(n3)−⋯+(−1)n−1(nn))a1+(−(n2)+(n3)−⋯+(−1)n−1(nn))a2+⋯              +((−1)n−1(nn))an\displaystyle = \left(\binom{n}{1} - \binom{n}{2} + \binom{n}{3} -\cdots +{(-1)}^{n-1}\binom{n}{n}\right) a_1 +\left(- \binom{n}{2} + \binom{n}{3} -\cdots +{(-1)}^{n-1}\binom{n}{n}\right) a_2 + \cdots \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \displaystyle + \left( {(-1)}^{n-1} \binom{n}{n}\right) a_n

Using the fact that LaTeX: −(n0)+(n1)−(n2)+(n3)−⋯+(−1)n−1(nn)=0-\binom{n}{0} +\binom{n}{1} - \binom{n}{2} + \binom{n}{3} -\cdots +{(-1)}^{n-1}\binom{n}{n} = 0,

=(n0)a1+((n0)−(n1))a2+⋯+((n0)−(n1)+(n2)−(n3)+⋯+(−1)n−1(nn−1))an\displaystyle = \binom{n}{0} a_1 + \left(\binom{n}{0} - \binom{n}{1}\right) a_2 + \cdots + \left(\binom{n}{0} -\binom{n}{1} + \binom{n}{2} - \binom{n}{3} +\cdots +{(-1)}^{n-1}\binom{n}{n-1}\right) a_n

Proposition #2 :∑r=0k(nr)(−1)r=(n−1k)(−1)k\displaystyle \text{Proposition \#2 :} \sum_{r=0}^k {\binom{n}{r} {(-1)}^r} = \binom{n-1}{k} {(-1)}^k

Proof:\text{Proof:}

We use induction (since it is very easy to observe using Pascal's triangle addition law).

S(k=0):(n0)=(n−10)\displaystyle S(k=0) : \binom{n}{0} = \binom{n-1}{0} . LaTeX: True\text{True}

S(k):∑r=0k(nr)(−1)r=∑r=0k−1(nr)(−1)r+(nk)(−1)k\displaystyle S(k) : \sum_{r=0}^k {\binom{n}{r} {(-1)}^r} = \sum_{r=0}^{k-1} {\binom{n}{r} {(-1)}^r} + \binom{n}{k} {(-1)}^k

=(n−1k−1)(−1)k−1+(nk)(−1)k=(n−1k)(−1)k\displaystyle = \binom{n-1}{k-1} {(-1)}^{k-1} + \binom{n}{k} {(-1)}^k = \binom{n-1}{k} {(-1)}^k

Q.E.D.■\displaystyle \mathbf{Q.E.D.}\blacksquare

Coming back to our proof, we use this formula.

=(n−10)a1−(n−11)a2+⋯−1n−1(n−1n−1)an\displaystyle = \binom{n-1}{0} a_1 - \binom{n-1}{1} a_2 + \cdots {-1}^{n-1} \binom{n-1}{n-1} a_n

=∑m=0n−1(n−1m)am+1(−1)m\displaystyle = \sum_{m=0}^{n-1}{\binom{n-1}{m} a_{m+1} {(-1)}^m}

Q.E.D.■\displaystyle \mathbf{Q.E.D.}\blacksquare

Now if we substitute ak=1ka_k = \frac{1}{k},

∑m=1n(nm)Hm(−1)m−1=∑m=0n−1(n−1m)(−1)mm+1=1n\displaystyle \sum_{m=1}^n {\binom{n}{m} H_m {(-1)}^{m-1}} = \sum_{m=0}^{n-1}{\binom{n-1}{m} \frac{{(-1)}^m}{m+1}} = \frac{1}{n}

Note that HmH_m denotes the harmonic number which is given by

Hm=∑k=1m1mH_m = \displaystyle \sum_{k=1}^m \dfrac 1m

as stated in your problem.

Tapas Mazumdar - 4 years, 2 months ago
×

Problem Loading...

Note Loading...

Set Loading...