JEE Main 2022 — Binomial Theorem Question with Solution

Given, Binomial expansion

$${\left(2x^{\frac{1}{5}}-\frac{1}{x^{\frac{1}{5}}}\right)}^{15}$$

$$\therefore$$ General Term

$$T_{r+1}={}^{15}{C}_r.{\left(2x^{\frac{1}{5}}\right)}^{15-r}.{\left(-\frac{1}{x^{\frac{1}{5}}}\right)}^r$$

$$={}^{15}{C}_r.2^{15-r}.x^{\frac{1}{5}(15-r-r)}.(-1{)}^r$$

For $$x^{-1}$$ term;

$$\frac{1}{5}(15-2r)=-1$$

$$\Rightarrow 15-2r=-5$$

$$\Rightarrow 2r=20$$

$$\Rightarrow r=10$$

m is the coefficient of $$x^{-1}$$ term,

$$\therefore$$ $$m={}^{15}{C}_{10}.2^{15-10}.(-1{)}^{10}$$

$$={}^{15}{C}_{10}.2^5$$

For $$x^{-3}$$ term;

$$\frac{1}{5}(15-2r)=-3$$

$$\Rightarrow 15-2r=-15$$

$$\Rightarrow 2r=30$$

$$\Rightarrow r=15$$

n is the coefficient of $$x^{-3}$$ term,

$$\therefore$$ $$n={}^{15}{C}_{15}.2^{15-15}.(-1{)}^{15}$$

$$=1.1.-1$$

$$=-1$$

Given,

$$m{n}^2={}^{15}{C}_r.2^r$$

$$\Rightarrow {}^{15}{C}_{10}.2^5.(1{)}^2={}^{15}{C}_r.2^r$$ [putting value of m and n]

$$\Rightarrow {}^{15}{C}_{15-10}.2^5={}^{15}{C}_r.2^r$$

$$\Rightarrow {}^{15}{C}_5.2^5={}^{15}{C}_r.2^r$$

Comparing both side, we get

$$r=5$$.