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$$\dfrac{\sum_{r=0}^{24} \binom{100}{4r}\binom{100}{4r+2}}{\sum_{r=1}^{25}\binom{200}{8r-6}}$$

The numerator is fine by multiplying 2 sequences we can easily get it but the denominator is taking a long time to be figured out using 8th roots of unity . I am searching for an alternate method to figure out the denominator .Please help

Alex
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1 Answers1

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Define $$D:= \sum_{r=1}^{25}\binom{200}{8r-6} = \sum_{r=0}^{24}\binom{200}{8r+2}$$ Then first note that by reindexing and using $\binom{n}{k} = \binom{n}{n-k}$, we have $$\sum_{r=0}^{24}\binom{200}{8r+2} = \sum_{r=0}^{24}\binom{200}{8(24-r)+2} = \sum_{r=0}^{24} \binom{200}{200-(8(24-r)+2)} = \sum_{r=0}^{24}\binom{200}{8r+6}$$

In particular, we note $$2^{199} = \sum_{k=0}^{100}\binom{200}{2r} = 2D+\sum_{r=0}^{50}\binom{200}{4r},$$ so $$\begin{align*}D &= 2^{198}-\frac{1}{2}\sum_{r=0}^{50}\binom{200}{4r} \\&= 2^{198}-\frac{1}{8}(2^{200}+(1+i)^{200}+(1-i)^{200}) \\ &= 2^{197}-2^{98}\end{align*}$$

Brian Moehring
  • 21,088
  • Yes but we need to find the ratio of N/D. Numerator comes out to be 200C98 so how can we find the ratio here. I guess we need to rearrange somehow. The answer given is 1 . – Alex Feb 18 '20 at 07:42
  • @Saddy No matter how I interpret your question, I can't get the answer to be 1. Do you want to double-check the entire problem? Alternatively, if this is from a textbook, which book and problem is it? – Brian Moehring Feb 18 '20 at 19:24
  • Yes it's a question by our school math teacher as a part of a worksheet. In my answersheet I had shown all the steps but they didn't give me full marks because I couldn't show the answer to be 1. – Alex Feb 19 '20 at 07:31
  • Could you please explain how could you get the 5th line where you found an equation for denominator? – Alex Feb 19 '20 at 07:46
  • Assuming you're talking about $$\sum_{r=0}^{50}\binom{200}{4r} = \frac{1}{4}(2^{200} + (1+i)^{200} + (1-i)^{200}),$$ it just comes from adding $$(1+x)^{200} = \sum_{k=0}^{200}\binom{200}{k}x^k$$ for $x=1, -1, i, -i$ and noting that $$1^k + (-1)^k + i^k + (-i)^k = \begin{cases}4 & \text{ if } 4 \mid k \ 0 & \text{ otherwise}\end{cases}$$ – Brian Moehring Feb 19 '20 at 10:05
  • Thanks a lot for the explanation – Alex May 20 '20 at 14:46