Combinatorial series

C(1000,50) + 2 * C(999,49)+3 * C(998,48) + 4 * C(997,47) ............................+ 51 * C(950,0) = S

S = C(n,k)

Find n & k .

Note by Purvam Modi
4 years, 6 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

We may use convolution of generating functions:

The g.f. of the sequence \(\langle \binom{950}{0}, \binom{951}{1}, \binom{952}{2} \cdots\rangle\) is

\(\displaystyle G(z)=\frac{1}{(1-z)^{951}}\)

The g.f. of the sequence \(\langle 1, 2, 3 \cdots\rangle\) is

\(\displaystyle H(z)=\frac{1}{(1-z)^{2}}\)

\(\displaystyle \therefore G(z)\cdot H(z) = \frac{1}{(1-z)^{953}} = \sum_{n\ge 0}\left(\sum_{k=0}^n (n-k+1)\, \binom{950+k}{k}\right) z^n\)

We require \([z^{50}]\), which is \(\binom{1002}{50}\)

Hence, n=1002 and k=50 or 952

Gopinath No - 4 years, 6 months ago

Log in to reply

Your answer is exactly right and thanks for the solution !

Purvam Modi - 4 years, 6 months ago

Log in to reply

I tried to find that but I used partial fractions. Unfortunately, I went up with S = 44314698 - 48452 ((1/50!) + (1/49!) + (1/48!) + ... + (1/1!) + (1/0!)). In which I found it very hard to sum even its reciprocal.. good solution by the way...

John Ashley Capellan - 4 years, 6 months ago

Log in to reply

Okay, but isn't the S you came up with much smaller than the answer?

Gopinath No - 4 years, 6 months ago

Log in to reply

@Gopinath No Probably so.. it's that small... I don't know if I used the right method but it sounded somewhat credible...

John Ashley Capellan - 4 years, 6 months ago

Log in to reply

@John Ashley Capellan That's fine, happens sometimes.

Gopinath No - 4 years, 6 months ago

Log in to reply

Excuse me but isn't the last term has multiplier of 51 if we continue the pattern?

John Ashley Capellan - 4 years, 6 months ago

Log in to reply

Sorry Its 51

Purvam Modi - 4 years, 6 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...