Question:
If 3 rd, 4th 5th and 6th terms in the expansion of $(x+a)^{n}$ be respectively $a, b, c$ and $d$, prove that $\frac{b^{2}-a c}{c^{2}-b d}=\frac{5 a}{3 c}$.
Solution:
We have:
$(x+a)^{n}$
The $3 \mathrm{rd}, 4$ th, 5 th and 6 th terms are ${ }^{n} C_{2} x^{n-2} a^{2},{ }^{n} C_{3} x^{n-3} a^{3},{ }^{n} C_{4} x^{n-4} a^{4}$ and ${ }^{n} C_{5} x^{n-5} a^{5}$, respectively.
Now,
${ }^{n} C_{2} x^{n-2} a^{2}=a$
${ }^{n} C_{3} x^{n-3} a^{3}=b$
${ }^{n} C_{4} x^{n-4} a^{4}=c$
${ }^{n} C_{5} x^{n-5} a^{5}=d$
$\mathrm{LHS}=\frac{b^{2}-a c}{c^{2}-b d}$
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