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General Term of Binomial Expansion

In the expans...

Question

In the expansion of $(a + b)^{n}$ , the ratio of the binomial coefficients of $2^{n d}$ and $3^{r d}$ terms is equal to the ratio of the binomial coefficients of $5^{t h}$ and $4^{t h}$ terms, then $n =$

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Solution

The correct option is C $5$
It is given that coefficients of $\frac{T_{2}}{T_{3}} = \frac{T_{5}}{T_{4}}$
Therefore, $\frac{^{n} C_{1}}{^{n} C_{2}} = \frac{^{n} C_{4}}{^{n} C_{3}}$

$\Rightarrow \frac{n}{\frac{n (n - 1)}{2!}} = \frac{\frac{(n) (n - 1) (n - 2) (n - 3)}{4!}}{\frac{n (n - 1) (n - 2)}{3!}}$

$\Rightarrow \frac{2}{n - 1} = \frac{n - 3}{4}$
$8 = n^{2} - 4 n + 3$
$n^{2} - 4 n - 5 = 0$
$(n - 5) (n + 1) = 0$
$n = 5$ and $n = - 1$
However, $n$ is a natural number.
Therefore $n = 5$

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