Let 0 r n.

Question

Let $0\le r\le n$. If ${}^{n+1}{C}_{r+1}:{}^{n}{C}_{r}:{}^{n-1}{C}_{r-1}=55:35:21$, then $2n+5r$ is equal to :

Accepted Answer

Given $0\le r\le n$. If $(\frac{n+1}{r+1}):(\frac{n}{r}):(\frac{n-1}{r-1})=55:35:21$, then we are to determine the value of $2n+5r$.

Write the given proportions involving binomial coefficients:

$\frac{n+1}{r+1}	imes (\frac{n}{r}):(\frac{n}{r}):\frac{r}{n}	imes (\frac{n}{r})=55:35:21$

Simplify the proportions:

$\frac{n+1}{r+1}=\frac{55}{35}\, \, \frac{n}{r}=\frac{35}{21}$

From the simplified ratios, we establish the following two equations:

$\frac{n+1}{r+1}=\frac{11}{7}\, \Rightarrow \, 7(n+1)=11(r+1)\, \Rightarrow \, 7n-11r=4\,$

$\frac{n}{r}=\frac{5}{3}\, \Rightarrow \, 3n=5r\, \Rightarrow \, 3n-5r=0\,$

Solve equations (1) and (2) simultaneously:

From equation (2), solve for $n$:

$3n=5r\, \Rightarrow \, n=\frac{5r}{3}$

Substitute $n$ into equation (1):

$7(\frac{5r}{3})-11r=4\, \Rightarrow \, \frac{35r}{3}-11r=4$

$\frac{35r-33r}{3}=4\, \Rightarrow \, \frac{2r}{3}=4\, \Rightarrow \, 2r=12\, \Rightarrow \, r=6$

Substituting $r$ back to find $n$:

$n=\frac{5	imes 6}{3}=10$

Compute $2n+5r$:

$2n+5r=2	imes 10+5	imes 6=20+30=50$

Thus, the value of $2n+5r$ is 50.

Solutions · 1