Question
In the binomial expansion of $${\left( {a - b} \right)^n},n \geqslant 5,$$ the sum of the $${5^{th}}$$ and $${6^{th}}$$ terms is zero. Then $$\frac{a}{b}$$ equals
A.
$$\frac{{\left( {n - 5} \right)}}{6}$$
B.
$$\frac{{\left( {n - 4} \right)}}{5}$$
C.
$$\frac{5}{{\left( {n - 4} \right)}}$$
D.
$$\frac{6}{{\left( {n - 5} \right)}}$$
Answer :
$$\frac{{\left( {n - 4} \right)}}{5}$$
Solution :
$${\left( {a - b} \right)^n},n \geqslant 5$$
In binomial expansion of above $${T_5} + {T_6} = 0$$
$$\eqalign{
& \Rightarrow \,{\,^n}{C_4}{a^{n - 4}}{b^4} + {\,^n}{C_5}{a^{n - 5}}{b^5} = 0 \cr
& \Rightarrow \,\,\frac{{^n{C_4}}}{{^n{C_5}}}.\frac{a}{b} = 1 \cr
& \Rightarrow \,\,\frac{{4 + 1}}{{n - 4}}.\frac{a}{b} = 1 \cr
& \Rightarrow \,\,\frac{a}{b} = \frac{{n - 4}}{5} \cr} $$