JEE Main 2022 — Binomial Theorem Question with Solution

Note : Multinomial Theorem :

The general term of $${\left(x_1+x_2+...+x_n\right)}^n$$ the expansion is

$$\frac{n!}{n_1!n_2!...n_n!}{x}_1^{n_1}{x}_2^{n_2}...x_n^{n_n}$$

where n₁ + n₂ + ..... + n_n = n

Given,

$${\left(3x^2-2x^2+\frac{5}{x^5}\right)}^{10}$$

$$=\frac{{(3x^8-2x^7+5)}^{10}}{x^{50}}$$

Now constant term in $${\left(3x^3-2x^2+\frac{5}{x^5}\right)}^{10}=x^{50}$$ term in $$(3x^8-2x^7+5{)}^{10}$$

General term in $$(3x^8-2x^7+5{)}^{10}$$ is

$$=\frac{10!}{n_1!n_2!n_3!}(3x^8{)}^{n_1}(-2x^7{)}^{n_2}(5{)}^{n_3}$$

$$=\frac{10!}{n_1!n_2!n_3!}(3{)}^{n^1}(-2{)}^{n_2}(5{)}^{n^3}.x^{8n_1+7n_2}$$

$$\therefore$$ Coefficient of $$x^{8n_1+7n_2}$$ is

$$=\frac{10!}{n_1!n_2!n_3!}(3{)}^{n_1}(-2{)}^{n_2}(5{)}^{n_3}$$

where $$n_1+n_2+n_3=0$$

For coefficient of x⁵⁰ :

$$8n_1+7n_2=50$$

$$\therefore$$ Possible values of n₁, n₂ and n₃ are

n$${}_1$$	n$${}_2$$	n$${}_3$$
1	6	3

$$\therefore$$ Coefficient of x⁵⁰

$$=\frac{10!}{1!6!3!}(3{)}^1(-2{)}^6(5{)}^3$$

$$=\frac{10\times 9\times 8\times 7}{6}\times 3\times 5^3\times 2^6$$

$$=5\times 3\times 8\times 7\times 3\times 5^3\times 2^6$$

$$=7\times 5^4\times 3^2\times 2^9$$

$$=2^k.l$$

$$\therefore$$ $$l=7\times 5^4\times 3^2$$ = An odd integer

and $$2^k=2^9$$

$$\Rightarrow k=9$$