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Binomial Theorem

Let tn deno...

Question

Let $t_{n}$ denote the $n t h$ term in a binomial expansion. If $\frac{t_{6}}{t_{5}}$ in the expansion of $(a + b)^{n + 4}$ and $\frac{t_{5}}{t_{4}}$ in the expansion of $(a + b)^{n}$ are equal, then $n$ is equal to

A

9

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B

11

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C

13

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D

15

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E

17

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Solution

The correct option is D $15$
$t_{6}$ and $t_{5}$ in the expansion of $(a + b)^{n + 4}$ is

$t_{5} = t_{4 + 1} =^{n + 4} C_{4} a^{n + 4 - 4} b^{4} =^{n + 4} C_{4} a^{n} \cdot b^{4}$
and $t_{6} = t_{5 + 1} =^{n + 4} C_{5} a^{n + 4 - 5} b^{5} =^{n + 4} C_{5} a^{n - 1} b^{5}$
Therefore, $\frac{t_{6}}{t_{5}} = \frac{^{n + 4} C_{5} \cdot a^{n - 1} \cdot b^{5}}{^{n + 4} C_{4} \cdot a^{n} \cdot b^{4}} = \frac{^{n + 4} C_{5}}{^{n + 4} C_{4}} (\frac{b}{a}) = \frac{n}{5} (\frac{b}{a})$ .... (i)
Now, $t_{5}$ and $t_{4}$ in the expansion of $(a + b)^{n}$ is
$t_{5} = t_{4 + 1} =^{n} C_{4} \cdot a^{n - 4} \cdot b^{4}$
and $t_{4} = t_{3 + 1} =^{n} C_{3} \cdot a^{n - 3} \cdot b^{3}$
Therefore, $\frac{t_{5}}{t_{4}} = \frac{^{n} C_{4} \cdot a^{n - 4} \cdot b^{4}}{^{n} C_{3} \cdot a^{n - 3} \cdot b^{3}} = \frac{n - 3}{4} \cdot (\frac{b}{a})$ .... (ii)
On equating Eqs. (i) and (ii), we get
$\frac{n}{5} (\frac{b}{a}) = \frac{n - 3}{4} (\frac{b}{a})$
$\Rightarrow 4 n = 5 n - 15$

$\Rightarrow n = 15$

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\frac{T_{2}}{T_{3}}

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