JEE Mathematics 2023 Question with Solution

Given: We need the constant term in the expansion of $\left(2x + \frac{1}{x^7} + 3x^2\right)^5$.

Find: The numerical value of the constant term.

Using the multinomial theorem, a general term is

$$T = \frac{5!}{r_1! \, r_2! \, r_3!}(2x)^{r_1}\left(\frac{1}{x^7}\right)^{r_2}(3x^2)^{r_3}$$

where

$$r_1 + r_2 + r_3 = 5$$

For the constant term, the power of $x$ must be zero. Hence,

$$r_1 - 7r_2 + 2r_3 = 0$$

So we solve

$$\begin{aligned} r_1 + r_2 + r_3 &= 5 \\ r_1 - 7r_2 + 2r_3 &= 0 \end{aligned}$$

From the working, the only possibility is

$$r_1 = 1, \quad r_2 = 1, \quad r_3 = 3$$

Now the constant term is

$$\frac{5!}{1! \, 1! \, 3!}(2)^1(3)^3$$ $$= 20 \times 2 \times 27$$ $$= 1080$$

Therefore, the constant term is $1080$.

Given: The expression is $\left(2x + \frac{1}{x^7} + 3x^2\right)^5$.

Find: The constant term.

Let the powers chosen from the three terms be $n_1, n_2, n_3$ respectively. Then

$$n_1 + n_2 + n_3 = 5$$

and for the constant term,

$$n_1 + 2n_3 = 7n_2$$

Checking non-negative integer solutions consistent with the total power $5$ gives

$$n_1 = 1, \quad n_2 = 1, \quad n_3 = 3$$

Hence the required term is

$$\frac{5!}{1! \, 1! \, 3!}(2x)^1\left(\frac{1}{x^7}\right)^1(3x^2)^3$$

The power of $x$ becomes

$$1 - 7 + 6 = 0$$

so this is indeed the constant term. Its coefficient is

$$\frac{5!}{1! \, 1! \, 3!} \cdot 2 \cdot 3^3 = 20 \cdot 2 \cdot 27 = 1080$$

Therefore, the constant term is $1080$.

Note: one provided approach incorrectly substitutes $r_1 = 3, r_2 = 1, r_3 = 1$ in text, but the coefficient computation and final answer correspond to the correct values $1,1,3$.