Solveeit Logo
Login

Question

Mathematics Question on Binomial theorem

In the binomial expansion of (a−b)n,  n  ≥ 5,{{(a-b)}^{n}},\,\,n\,\,\ge \,5, the sum of 5th5^{th} and 6th6^{th} term is zero. Then, ab\frac{a}{b} is equal to

A

n−42\frac{n-4}{2}

B

n−43\frac{n-4}{3}

C

n−45\frac{n-4}{5}

D

n−44\frac{n-4}{4}

Answer

n−45\frac{n-4}{5}

Explanation

Solution

Given expansion is (a−b)n.{{(a-b)}^{n}}.
∴\therefore T5=nC4(a)(n−4)(−b)4{{T}_{5}}{{=}^{n}}{{C}_{4}}{{(a)}^{(n-4)}}{{(-b)}^{4}}
and T6=nC5(a)(n−5)(−b)5{{T}_{6}}{{=}^{n}}{{C}_{5}}{{(a)}^{(n-5)}}{{(-b)}^{5}}
According to the given condition,
T5+T6=0{{T}_{5}}+{{T}_{6}}=0
∴\therefore nC4(a)n−4(−b)4+nC5(a)n−5(−b)5=0^{n}{{C}_{4}}{{(a)}^{n-4}}{{(-b)}^{4}}{{+}^{n}}{{C}_{5}}{{(a)}^{n-5}}{{(-b)}^{5}}=0
⇒\Rightarrow (an−4)(−b)4[nC4+nC5(−ba)]=0({{a}^{n-4}}){{(-b)}^{4}}\left[ ^{n}{{C}_{4}}{{+}^{n}}{{C}_{5}}\left( \frac{-b}{a} \right) \right]=0
⇒\Rightarrow ab=nC5nC4=n(n−1) (n−2) (n−3) (n−4)5×4×3×2×1n(n−1) (n−2) (n−3)4×3×2×1\frac{a}{b}=\frac{^{n}{{C}_{5}}}{^{n}{{C}_{4}}}=\frac{\frac{n(n-1)\,(n-2)\,(n-3)\,(n-4)}{5\times 4\times 3\times 2\times 1}}{\frac{n(n-1)\,(n-2)\,(n-3)}{4\times 3\times 2\times 1}}
=(n−4)5=\frac{(n-4)}{5}