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Find three consecutive entries of a row of Pascal triangle that are in the ratio of 1 : 2 : 3

This means that:

$$\begin{align} 2\binom{n}{k} =\binom{n}{k+1}\\ 3\binom{n}{k} =\binom{n}{k+2}\\ \end{align}$$

I simplefied these equations:

$$\begin{align} 2\binom{n}{k} =\binom{n}{k+1}\\ \end{align}$$

Is the same as:

$$\begin{align} 2k+2 = n-k\\ 3k+2 = n \end{align}$$

And:

$$\begin{align} 3\binom{n}{k} =\binom{n}{k+2}\\ \end{align}$$

Is the same as:

$$\begin{align} 3(k+1)(k+2) = (n-k-1)(n-k) \end{align}$$

I do not know how I could go further so that I would end up with the value of n and k. This is because I can not simplify 3(k+1)(k+2) = (n-k-1)(n-k) enough.

How can I proceed so that I can get the value of n and k for:

$$\begin{align} 3(k+1)(k+2) = (n-k-1)(n-k) \\ 3k+2 = n \end{align}$$

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  • $\begingroup$ It's easier if you write it $3 \binom{n}{k+1} = 2\binom{n}{k+2}$. Then you get another linear relation between $n$ and $k$. $\endgroup$ – Daniel Fischer Oct 24 '15 at 10:26
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Since $n=3k+2$, you can have $$3(k+1)(k+2)=(\color{red}{3k+2}-k-1)(\color{red}{3k+2}-k)$$ Now solve this for $k$. (don't forget to check if each $k$ is sufficient.)

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Alternatively, coming from this question, dividing the consecutive binomial coefficients yields $$\frac21=\frac{\tbinom{n}{k+1}}{\tbinom nk}=\frac{n-k}{k+1}\qquad\text{ and }\qquad \frac32=\frac{\tbinom{n}{k+2}}{\tbinom{n}{k+1}}=\frac{n-k-1}{k+2},$$ and solving both for $n$ yields $n=3k+2$ and $n=\tfrac52k+4$, respectively, so $k=4$ and $n=14$.

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