Mathematics - Binomial Theorem Question with Solution | TestHub

General term$=\frac{10!}{r_1!r_2!r_3!}{\left(2x^2\right)}^{r_1}{\left(3x\right)}^{r_2}{\left(4\right)}^{r_3}$

$a_7=\frac{10!.2^3.3{.4}^6}{3!1!6!}+\frac{10!2^2.3^3.4^5}{2!3!5!}+\frac{10!.2{.3}^5.4^4}{1!.5!.4!}+\frac{10!.3^7.4^3}{7!3!}$

$a_7=2^3\left[\frac{10!.2^6.3{.4}^3}{6!1!3!}+\frac{10!2^5.3^3.4^2}{5!3!2!}+\frac{10!.2^4.3^5.4^1}{4!.5!.1!}+\frac{10!2^3.3^7.4^0}{3!7!}\right]$

$a_{13}=\frac{10!.2^6.3{.4}^3}{6!1!3!}+\frac{10!2^5.3^3.4^2}{5!3!2!}+\frac{10!.2^4.3^5.4^1}{4!.5!.1!}+\frac{10!2^3.3^7.4^0}{3!7!}$

$\frac{a_7}{a_{13}}=2^3\left[\frac{\frac{10!.3{.2}^{12}}{3!1!6!}+\frac{10!.3^3.2^9}{2!.3!.5!}+\frac{10!.3^5.2^6}{1!.5!.4!}+\frac{10!.3^7.2^3}{7!.3!}}{\frac{10!.3{.2}^{12}}{3!.1!.6!}+\frac{10!.3^3.2^9}{2!.3!.5!}+\frac{10!.3^5.2^6}{1!.5!.4!}+\frac{10!.3^7.2^3}{7!.3!}}\right]=2^3=8$