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sum of expression $$\large\frac{\sum^{r}_{k=0}\binom{n}{k}\binom{n-2k}{r-k}}{\sum^{n}_{k=r}\binom{2k}{2r}\bigg(\frac{3}{4}\bigg)^{n-k}\bigg(\frac{1}{2}\bigg)^{2k-2r}}(n\geq 2r)$$

$(a)\;1/2\;\;\;\;\;\; (b)\;2\;\;\;\;\;\; (c)\; 1\;\;\;\;\;\; (d)\;$ none

what i try

i put $r=1,n=2$

$$\Large \frac{\sum^{1}_{k=0}\binom{2}{k}\binom{2-2k}{1-k}}{\sum^{2}_{k=1}\binom{2k}{2}\bigg(\frac{3}{4}\bigg)^{2-k}\bigg(\frac{1}{2}\bigg)^{2k-2}}$$

$$\frac{\binom{2}{0}\binom{2}{1}+\binom{2}{1}\binom{0}{0}}{\binom{2}{2}\frac{3}{4}+\binom{4}{2}\frac{1}{4}} = \frac{16}{9}$$

How do i solve it help me please

jacky
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  • Then the answer is: d) – user52227 Feb 06 '19 at 12:51
  • can you show how you get it @user52227 – jacky Feb 06 '19 at 13:07
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    The numerator expression has a very unpredictable closed form. For $r = 2$ I checked it is equal to $n(2n-3)$, and then for $n=3$ I checked it is equal to $\frac{4n^3 - 18n^2 + 17n}{3}$ and then for $n=4$ I checked it is $\frac{4n^4 - 36n^2 + 95n^2 - 69n}{6}$. I'd like you to figure out a pattern if you can! I think there has been a printing mistake, or it is a trick question : there is no way you were expected to evaluate that unless it were easy or the answer was to just find $r,n$ and ensure the answer was none. – Sarvesh Ravichandran Iyer Feb 07 '19 at 12:20

2 Answers2

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Your own choice of the numbers yield the required answer i,e, none.

user642005
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  • because the answers, as given, are independent of $r$ and $n$, if you plug any values for them (within those allowed) and get a different value from a), b), c) then it is none of them – G Cab Feb 06 '19 at 13:51
  • @user642005 i know that but How do i solve it . explain me please also G Cab can you please explain me – jacky Feb 06 '19 at 16:52
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A detailed canonical answer is required

A detailed canonical answer is the following.

Given a number $x$ by $P(x)$ we denote the following claim:

For each $n\geq 2r$, $x$ is the value of the expression $$\large\frac{\sum^{r}_{k=0}\binom{n}{k}\binom{n-2k}{r-k}}{\sum^{n}_{k=r}\binom{2k}{2r}\bigg(\frac{3}{4}\bigg)^{n-k}\bigg(\frac{1}{2}\bigg)^{2k-2r}}.$$

I guess you were asked which of the claims $P(1/2)$, $P(2)$, $P(1)$, and $N=\neg(P(1/2)\vee P(2)\vee P(1))$ holds. Remark that variables $n$ and $r$ are under the quantors $\forall$ in the formalizations of the claims $P(x)$ and the claim $P(x)$ is false provided there exist $(n\geq 2r)$ such that the expression does not equal $P(x)$. In particular, to show that $P(x)$ is false we don’t need to calculate the value of the expression for all $n\geq 2r$ or to find its closed form. Since for $r=1$, $n=2$ the expression equals $\frac {16}{9}$, we see that each of the claims $P(1/2)$, $P(2)$, and $P(1)$, is false. Therefore the claim $P=P(1/2)\vee P(2)\vee P(1)$ is false too, so the claim $\neg P$ (that is $N$) is true.

Alex Ravsky
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