Binomial Coefficient Challenge 2!

\[\displaystyle \sum_{k=0}^n{\binom{n+k}{2k}{(-4)}^k} = {(-1)}^{n}(2n+1)\]

Note by Kartik Sharma
12 months 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

Denote the LHS as \(A_{n}\). Using the identity \(\dbinom{n}{r} + \dbinom{n}{r-1} = \dbinom{n+1}{r} \) and re-indexing wherever required, it is easy to show that,

\[A_{n+1} + A_{n-1} + 2A_{n} = 0\]

Using the initial values \(A_{0} = 1\), \(A_{1} = -3\), it follows that \(A_{n} = (-1)^n (2n+1)\).

Ishan Singh - 11 months, 3 weeks ago

Log in to reply

Nice. Probably you would like this.

Kartik Sharma - 11 months, 3 weeks ago

Log in to reply

The Fibonacci polynomials \(F_n(x)\) are defined by \[ F_{n+1}(x) \; = \; xF_n(x) + F_{n-1}(x) \hspace{1cm} n \ge 1 \hspace{3cm} F_0(x) \; = \;0\,,\,F_1(x) \; = \; 1 \] They have generating function \[ G(x,t) \; = \; \sum_{n \ge 0} F_n(x)t^n \; = \; \frac{t}{1 - xt - t^2} \] Expanding this generating function we obtain, in particular, that \[ F_{2n+1}(x) \; = \; \sum_{j=0}^n {n+j \choose n-j} x^{2j} \] and hence the question asks us to evaluate \(F_{2n+1}(2i)\). Now \[ G(2i,t) \; = \; \frac{t}{1 - 2it - t^2} \; = \; \frac{t}{(1 - it)^2} \; = \; \sum_{m \ge 0} (m+1)i^m t^{m+1} \] so that \[ F_{2n+1}(2i) \; = \; (2n+1)i^{2n} \; = \; (-1)^n (2n+1) \]

Mark Hennings - 11 months, 3 weeks ago

Log in to reply

Nice! In fact this can be very easily converted into Gauss hypergeometric.

Kartik Sharma - 11 months, 3 weeks ago

Log in to reply

Sure, and we can do the third one by replacing \(\frac{(-1)^k}{k+1}\) by \(x^k\), identifying the new series as a \({}_3F_2\) hypergeometric at \(x\), and integrating that result to obtain the desired series as the value of a \({}_4F_3\) terminating hypergeometric at unity, and then throwing a bundle of hypergeometric identities at the problem. It would be nice to find a combinatoric proof instead.

Mark Hennings - 11 months, 3 weeks ago

Log in to reply

@Mark Hennings Yes. I agree, that's why I posted them here. BTW, there is a better way to get the third one.

Kartik Sharma - 11 months, 3 weeks ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...