Question

The constant term in the expansion of $\left(2 x+\frac{1}{x^7}+3 x^2\right)^5$ is_____

Answer 1

Correct Answer: 1080

The correct answer is 1080.
General term is $\sum \frac{5!(2x{)}^{n_1}{\left(x^{-7}\right)}^{n_2}{\left(3x^2\right)}^{n_3}}{n_1!n_2!n_3!}$
For constant term,
$n_1+2n_3=7n_2$
$\&n_1+n_2+n_3=5$
Only possibility $n_1=1,n_2=1,n_3=3$
$\Rightarrow$ constant term $=1080$

Answer 2

1. General Term in the Expansion: Use the multinomial theorem: \[ T = \frac{5!}{r_1! \, r_2! \, r_3!} \cdot (2x)^{r_1} \cdot \left(\frac{1}{x^7}\right)^{r_2} \cdot (3x^2)^{r_3}, \] where \( r_1 + r_2 + r_3 = 5 \). 2. Condition for Constant Term: The net power of \( x \) in \( T \) is: \[ r_1 - 7r_2 + 2r_3 = 0. \] Solve for \( r_1, r_2, r_3 \) under \( r_1 + r_2 + r_3 = 5 \). 3. Solve for \( r_1, r_2, r_3 \): Using the conditions: \[ r_1 + r_2 + r_3 = 5, \quad r_1 - 7r_2 + 2r_3 = 0. \] Substitute \( r_1 = 3 \), \( r_2 = 1 \), \( r_3 = 1 \). 4. Compute the Coefficient: The constant term is: \[ T = \frac{5!}{3! \, 1! \, 1!} \cdot (2)^3 \cdot \left(\frac{1}{x^7}\right)^1 \cdot (3)^1 = 10 \cdot 8 \cdot 3 = 1080. \]