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# 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
3 years, 7 months ago

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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 · 3 years, 7 months ago

Your answer is exactly right and thanks for the solution ! · 3 years, 7 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... · 3 years, 7 months ago

Okay, but isn't the S you came up with much smaller than the answer? · 3 years, 7 months ago

Probably so.. it's that small... I don't know if I used the right method but it sounded somewhat credible... · 3 years, 7 months ago

That's fine, happens sometimes. · 3 years, 7 months ago

Excuse me but isn't the last term has multiplier of 51 if we continue the pattern? · 3 years, 7 months ago