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Sum of Coefficients of All Terms

In the binomi...

Question

In the binomial expansion of $(a + b)^{n}$ , the sum of $5^{t h} a n d 6^{t h}$ terms is zero, then $\frac{a}{b}$ is equal to

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Solution

The correct option is B

$T_{5} + T_{6} = 0$
$\Rightarrow \frac{T_{6}}{T_{5}} = - 1$
$T_{5} =^{n} C_{4} a^{n - 4} b^{4}$
$T_{6} =^{n} C_{5} a^{n - 5} b^{5}$
$\Rightarrow \frac{T_{6}}{T_{5}} = \frac{^{n} C_{5} a^{n - 5} b^{5}}{^{n} C_{4} a^{n - 4} b^{4}}$
$\Rightarrow \frac{T_{6}}{T_{5}} = \frac{^{n} C_{5}}{^{n} C_{4}} \frac{b}{a}$
$\Rightarrow \frac{T_{6}}{T_{5}} = \frac{n - 5 + 1}{5} \frac{b}{a}$
$\Rightarrow \frac{T_{6}}{T_{5}} = \frac{n - 4}{5} \frac{b}{a} = - 1$
$\Rightarrow \frac{a}{b} = - (\frac{n - 4}{5})$

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