Binomial Theorem 1 Question 36
38. The coefficients of three consecutive terms of $(1+x)^{n+5}$ are in the ratio $5: 10: 14$. Then, $n$ is equal to (2013 Adv.)
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Solution:
- Let the three consecutive terms in $(1+x)^{n+5}$ be $t _r, t _{r+1}, t _{r+2}$ having coefficients
${ }^{n+5} C _{r-1},{ }^{n+5} C _r,{ }^{n+5} C _{r+1}$.
Given, ${ }^{n+5} C _{r-1}:{ }^{n+5} C _r:{ }^{n+5} C _{r+1}=5: 10: 14$
$\therefore \quad \frac{{ }^{n+5} C _r}{{ }^{n+5} C _{r-1}}=\frac{10}{5}$ and $\frac{{ }^{n+5} C _{r+1}}{{ }^{n+5} C _r}=\frac{14}{10}$
$\Rightarrow \quad \frac{n+5-(r-1)}{r}=2$ and $\frac{n-r+5}{r+1}=\frac{7}{5}$
$\Rightarrow \quad n-r+6=2 r$ and $5 n-5 r+25=7 r+7$
$\Rightarrow \quad n+6=3 r$ and $5 n+18=12 r$
$\therefore \quad \frac{n+6}{3}=\frac{5 n+18}{12}$
$\Rightarrow \quad 4 n+24=5 n+18 \Rightarrow \quad n=6$
