JEE Mathematics 2023 Question with Solution

Given:

$$\left(x - \frac{3}{x^2}\right)^n$$

The sum of the coefficients of the first three terms is $376$.

Find: The coefficient of $x^4$.

For the binomial expansion, the first three terms correspond to $r=0,1,2$. Their coefficients are:

$${}^nC_0, \quad -3\,{}^nC_1, \quad 3^2\,{}^nC_2$$

So,

$${}^nC_0 - 3\,{}^nC_1 + 3^2\,{}^nC_2 = 376$$

Using ${}^nC_0=1$, ${}^nC_1=n$ and ${}^nC_2=\frac{n(n-1)}{2}$,

$$1 - 3n + 9\cdot \frac{n(n-1)}{2} = 376$$

This simplifies to

$$3n^2 - 5n - 250 = 0$$

Factoring,

$$(n-10)(3n+25)=0$$

Since $n \in \mathbb{N}$,

$$n=10$$

Now the general term is

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

For the coefficient of $x^4$,

$$10-3r=4$$ $$r=2$$

Hence the required coefficient is

$${}^{10}C_2(-3)^2 = 45 \cdot 9 = 405$$

Therefore, the coefficient of $x^4$ is $405$.

Given:

$$\left(x - \frac{3}{x^2}\right)^n$$

with sum of coefficients of the first three terms equal to $376$.

Find: The coefficient of $x^4$.

The general term in the expansion is

$${}^nC_r\,x^{n-r}\left(-\frac{3}{x^2}\right)^r$$ $$= {}^nC_r(-3)^r x^{n-r-2r}$$ $$= {}^nC_r(-3)^r x^{n-3r}$$

Now for $r=0,1,2$, the first three terms are:

$$T_1 = {}^nC_0 x^n,$$ $$T_2 = {}^nC_1(-3)x^{n-3},$$ $$T_3 = {}^nC_2 9 x^{n-6}$$

So the sum of their coefficients is

$${}^nC_0 - 3{}^nC_1 + 9{}^nC_2 = 376$$ $$1 - 3n + 9\cdot \frac{n(n-1)}{2} = 376$$ $$2 - 6n + 9n(n-1) = 752$$ $$2 - 6n + 9n^2 - 9n = 752$$ $$9n^2 - 15n - 750 = 0$$ $$3n^2 - 5n - 250 = 0$$ $$(n-10)(3n+25)=0$$

Thus,

$$n=10$$

To get the coefficient of $x^4$, set the power equal to $4$:

$$n-3r=4$$ $$10-3r=4$$ $$r=2$$

Therefore the required coefficient is

$${}^{10}C_2(-3)^2$$ $$= 45 \times 9 = 405$$

So the answer is $405$.