### Binomial Theorem Question 4

#### Question 4 - 24 January - Shift 2

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

## Show Answer

#### Answer: 405

#### Solution:

#### Formula: Bionomial Theorem, Important Results (i) Coefficient of $x^{m}$ in the expansion of $(a x^p + \frac{b}{x^q})^n$

Given Binomial $(x-\frac{3}{x^{2}})^{n}, x \neq 0, n \in N$,

Sum of coefficients of first three terms

$ \begin{aligned} & { }^{n} C_0-{ }^{n} C_1 \cdot 3+{ }^{n} C_2 3^{2}=376 \\ & \Rightarrow 3 n^{2}-5 n-250=0 \\ & \Rightarrow(n-10)(3 n+25)=0 \\ & \Rightarrow n=10 \end{aligned} $

Now general term ${ }^{10} C_r x^{10-r}(\frac{-3}{x^{2}})^{r}$

$ \begin{aligned} & ={ }^{10} C_r x^{10-r}(-3)^{r} \cdot x^{-2 r} \\ & ={ }^{10} C_r(-3)^{r} \cdot x^{10-3 r} \end{aligned} $

Coefficient of $x^{4} \Rightarrow 10-3 r=4$

$\Rightarrow r=2$

${ }^{10} C_2(-3)^{2}=405$