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 .

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
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Comments

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

John Ashley Capellan - 7 years, 6 months ago

Sorry Its 51

Purvam Modi - 7 years, 6 months ago

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 - 7 years, 6 months ago

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 - 7 years, 6 months ago

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

gopinath no - 7 years, 6 months ago

@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 - 7 years, 6 months ago

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

gopinath no - 7 years, 6 months ago

Your answer is exactly right and thanks for the solution !

Purvam Modi - 7 years, 6 months ago

Math	Appears 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}`	$\boxed{123}$