7.Binomial Theorem
hard

ધારોકે $0 \leq r \leq n$. જો ${ }^{n+1} C_{r+1}:{ }^n C_r:{ }^{n-1} C_{r-1}=55: 35: 21$ હોય, તો $2 n+5 r=$.........

A

$60$

B

$62$

C

$50$

D

$55$

(JEE MAIN-2024)

Solution

$ \frac{{ }^{n+1} C_r}{{ }^n C_r}=\frac{55}{35} $

$ \frac{(n+1) !}{(r+1) !(n-r)} ! \frac{r !(n-r) !}{n !}=\frac{11}{7} $

$ \frac{(n+1)}{r+1}=\frac{11}{7}$

$ 7 \mathrm{n}=4+11 \mathrm{r} $

$ \frac{{ }^n C_r}{{ }^{n-1} C_{r-1}}=\frac{35}{21} $

$ \frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}=\frac{(\mathrm{r}-1) !(\mathrm{n}-\mathrm{r}) !}{(\mathrm{n}-1) !}=\frac{5}{3} $

$ \frac{\mathrm{n}}{\mathrm{r}}=\frac{5}{3} $

$ 3 \mathrm{n}=5 \mathrm{r} $

By solving $ \mathrm{r}=6 \quad \mathrm{n}=10 $

$ 2 \mathrm{n}+5 \mathrm{r}=50$

Standard 11
Mathematics

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