Browse Questions

# In the binomial expansion of $(a-b)^n,\:n\geq 5$ the sum of $5^{th}$ and $6^{th}$ term is $0$, then $\large\frac{a}{b}$=?

$\begin{array}{1 1} \frac{6}{n-5} \\ \frac{n-5}{6} \\ \frac{5}{n-4} \\ \frac{n-4}{5} \end{array}$

Toolbox:
• $T_{r+1}=^nC_r x^{n-r}.y^r$ in the expansion $(x+y)^n$
$T_5=^nC_4 a^{n-4}b^4$
$T_6=-^nC_5 a^{n-5}b^5$
Given $T_5+T_6=0$
$\Rightarrow\:^nC_4 a^{n-4}b^4-^nC_5 a^{n-5}b^5=0$
$\Rightarrow\:^nC_4 a^{n-4}b^4=^nC_5 a^{n-5}b^5$