Consider these two linear equation:
$$ 2a \binom{18}{j-1} - 4b \binom{18}{j} = \binom{18}{j-2}$$ True for $j=16,17$, find $a$ and $b$
This came up as part of a problem I was doing.. expanding everything out and doing it really ugly. Are there any tricks I could use to solve the system fast considering they are involving binomial coefficients? One thing I've considered doing is adding both equations and pascal's identity but not sure if that actually makes anything easier.
$$\frac{\binom{n}{r}}{\binom{n}{r-1}}=\frac{n+1-r}{r}$$ Thus, if you divide both side of the equation by $\binom{18}{j-1}$ , then you can simplify it and that will decrease your calculations.
For $j=16,17$ and using $\binom{18}{r}=\binom{18}{18-r}$, we have $$2a\binom{18}{3}-4b\binom{18}{2}=\binom{18}{4}$$ $$2a\binom{18}{2}-4b\binom{18}{1}=\binom{18}{3}$$
Writing out each binomial coefficient and cancelling common terms (eg, $18\cdot 17$ cancels off in first one), we have $$16a-6b=30 \tag{A}$$ $$51a-12b=136 \tag{B}$$
On $B-2A$ is obtained, $$a=4 , \quad b =17/3$$
Applying the identity : $$\frac{\binom{n}{r}}{\binom{n}{r-1}}=\frac{n+1-r}{r}$$
On
$$ 2a \binom{18}{j-1} - 4b \binom{18}{j} = \binom{18}{j-2}$$
First divide:
$$ 2a - 4b \frac{\binom{18}{j}}{\binom{18}{j-1}} = \frac{\binom{18}{j-2}}{\binom{18}{j-1}}$$
Apply the Id
$$2a - 4b (\frac{19-j}{j} )= (\frac{j-2}{19-j})$$
This is much better because, the solution's of the equation directly fall out of differencing the above expression over j=16 and j=17