Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Greatest term in the binomial expansion of $(a+2 x)^9$ when $a=1 \& x=\frac{1}{3}$ is :

Binomial Theorem

Solution:

$\frac{ T _{ r +1}}{ T _{ r }}=\frac{{ }^9 C _{ r } a ^{9- r } \cdot(2 x )^{ r }}{{ }^9 C _{ r -1} a ^{9- r +1}(2 x )^{ r -1}}=\frac{9- r +1}{ r } \frac{2 x }{ a }=\frac{9- r +1}{ r } \frac{2}{3}=\frac{10- r }{ r } \frac{2}{3}$
$\therefore T _{ r +1}> T _{ r }$ if $\frac{10- r }{ r } \frac{2}{3}>1$ i.e. $20>5 r$ i.e. $r <4$
Hence for values of ' $r$ ' upto 3, $T_{r+1} when $r =4, T _{ r +1}= T _{ r }$ i.e. $T _4= T _5 \&$ both of them are the greatest terms