JEE Main 2022 — Binomial Theorem Question with Solution

Given,

Binomial expression ${\left(2x^3+\frac{3}{x}\right)}^{10}$, so its $n^{th}$ term is given by,

$T_{r+1}={}^{10}C_r{\left(2x^3\right)}^{10-r}{\left(\frac{3}{x}\right)}^r$

$={}^{10}C_r{2}^{10-r}{3}^r{x}^{30-4r}$

Now putting $r=0,1,2,.....7$ and adding all term we get 

${\left(2x^3+\frac{3}{x}\right)}^{10}={}^{10}C_0{2}^{10}{3}^0{x}^{30}+{}^{10}C_1{2}^9{3}^1{x}^{26}..............{}^{10}C_7{2}^3{3}^7{x}^2+{}^{10}C_8{2}^2{3}^8\frac{1}{x^2}....$

${\left(2x^3+\frac{3}{x}\right)}^{10}-{}^{10}C_8{2}^2{3}^8\frac{1}{x^2}.......={}^{10}C_0{2}^{10}{3}^0{x}^{30}+{}^{10}C_1{2}^9{3}^1{x}^{26}..............{}^{10}C_7{2}^3{3}^7{x}^2$

Put $x=1$ both side we get,

$5^{10}-3^9\left(60+20+3\right)={}^{10}C_0{2}^{10}{3}^0+{}^{10}C_1{2}^9{3}^1..............{}^{10}C_7{2}^3{3}^7$

$5^{10}-3^9\left(83\right)=\text{sum of even coefficients}$

So $\beta =83$