Question

Let the sum of the coefficients of the first three terms in the expansion of $\left(x-\frac{3}{x^2}\right)^n, x \neq 0 n \in N$, be $376$. Then the coefficient of $x^4$ is ______

Answer 1

Correct Answer: 405

The correct answer is 405
Given Binomial ${\left(x-\frac{3}{x^2}\right)}^n,x\ne 0,n\in N$,
Sum of coefficients of first three terms
${}^n{C}_0-{}^n{C}_1\cdot 3+{}^n{C}_2{3}^2=376$
$\Rightarrow 3n^2-5n-250=0$
$\Rightarrow (n-10)(3n+25)=0$
$\Rightarrow n=10$
Now general term ${}^{10}{C}_r{x}^{10-r}{\left(\frac{-3}{x^2}\right)}^r$
$={}^{10}{C}_r{x}^{10-r}(-3{)}^r\cdot x^{-2r}$
$={}^{10}{C}_r(-3{)}^r\cdot x^{10-3r}$
Coefficient of $x^4\Rightarrow 10-3r=4$
$\Rightarrow r=2$
${}^{10}{C}_2(-3{)}^2=405$

Answer 2

Step 1: Expand the series 

The general term in the expansion of the series is given by:

\[ nC_r \cdot x^{n-r} \cdot (-3)^{2r} = nC_r \cdot (-3)^r \cdot x^{n - 3r}. \] 

Step 2: Sum of coefficients of the first three terms

The first three terms correspond to \(r = 0, 1, 2\). These terms are:

\[ T_0 = nC_0 \cdot x^n, \quad T_1 = nC_1 \cdot (-3) \cdot x^{n-3}, \quad T_2 = nC_2 \cdot 9 \cdot x^{n-6}. \]

The sum of the coefficients is:

\[ nC_0 - nC_1 \cdot 3 + nC_2 \cdot 9 = 376. \] 

Step 3: Solve for \(n\)

To solve for \(n\), simplify the equation:

\[ 1 - 3n + \frac{n(n - 1)}{2} \cdot 9 = 376. \] Simplifying further: \[ 9n^2 - 27n - 752 = 0. \]

Now, solve the quadratic equation:

\[ n = 10. \] 

Step 4: Coefficient of \(x^4\)

To find the coefficient of \(x^4\), we set \(n - 3r = 4\). This gives:

\[ r = \frac{n - 4}{3} = \frac{10 - 4}{3} = 2. \]

The coefficient is:

\[ nC_2 \cdot (-3)^2 = 10C_2 \cdot 9 = 45 \cdot 9 = 405. \]