If a, b, c and d in any binomial expansion be the 6th, 7th, 8th and 9th terms respectively, then prove that

Question:

If $a, b, c$ and $d$ in any binomial expansion be the 6 th, 7 th, 8 th and 9th terms respectively, then prove that $\frac{b^{2}-a c}{c^{2}-b d}=\frac{4 a}{3 c}$.

Solution:

Suppose the binomial expression is $(1+x)^{n}$.

Then, the 6 th, 7 th, 8 th and 9 th terms are ${ }^{n} C_{5} x^{5},{ }^{n} C_{6} x^{6},{ }^{n} C_{7} x^{7}$ and ${ }^{n} C_{8} x^{8}$, respectively.

Now, we have:

$\frac{{ }^{n} C_{6 x^{6}}}{{ }^{n} C_{5} x^{5}}=\frac{b}{a}, \frac{{ }^{n} C_{8 x^{8}}}{{ }^{n} C_{7} x^{7}}=\frac{d}{c}$ and $\frac{{ }^{n} C_{7 x} 7}{{ }^{n} C_{6} x^{6}}=\frac{c}{b}$

$\Rightarrow \frac{n-5}{6}=\frac{b}{a}$ and $\frac{n-6}{7}=\frac{c}{b}$

$\Rightarrow \frac{n-6 / 7}{n-5 / 6}=\frac{c / b}{b / a}$

$\Rightarrow \frac{6 n-36}{7 n-35}=\frac{c}{a}$

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