Mathematics - Binomial Theorem Question with Solution | TestHub

Given the expansion$(a+2b+4ab{)}^{10}$

$T_{r+1} =a^{10}·b^{10}{\left(\frac{1}{b}+\frac{2}{a}+4\right)}^{10}$

$T_{r+1}=a^{10}·b^{10} {{}^{10}C}_{r_1+r_2} {\left(\frac{1}{b}\right)}^{r_1}{\left(\frac{2}{a}\right)}^{r_2}{4}^{10-r_1-r_2}$

$T_{r+1}=a^{10}·b^{10}\frac{10!{\left(\frac{1}{b}\right)}^{r_1}{\left(\frac{2}{a}\right)}^{r_2}{4}^{10-r_1-r_2}}{r_1!r_2!\left(10-r_1-r_2\right)!}$

Putting, $r_1=2, r_2=3$

Coefficient of $a^7{b}^8$ is $\frac{10!·2^3·4^{10-2-3}}{2!.3!5!}$

$=\frac{2^{13}·10!}{2!.3!5!}$

$=2^{16}·315$

$K.2^{16}=315..2^{16}$

Hence, $K=315$