JEE Main 2023 — Mathematics Algebra

We know that general term in the expansion of ${(x-\frac{3}{{x}^{2}})}^{n}$(x−x23​)n is given by,

$${T}_{r+1}=Crn{x}^{n-r}{(\frac{-3}{{x}^{2}})}^{r}$$Tr+1​=Crnxn−r(x2−3​)r

$$={(-1)}^{r}\times Crn{3}^{r}{x}^{n-r-2r}$$=(−1)r×Crn3rxn−r−2r

$${T}_{r+1}={(-1)}^{r}\times Crn{3}^{r}{x}^{n-3r}...............(1)$$Tr+1​=(−1)r×Crn3rxn−3r...............(1)

So, ${T}_{1}={T}_{0+1}=C0n{3}^{0}{x}^{n}={x}^{n}$T1​=T0+1​=C0n30xn=xn

$${T}_{2}=(-1)\times C1n{3}^{1}{x}^{n-3}$$T2​=(−1)×C1n31xn−3

$${T}_{3}=C2n{3}^{2}{x}^{n-6}$$T3​=C2n32xn−6

Now given sum is

$$1-C1n\cdot 3+C2n\cdot {3}^{2}=376$$1−C1n⋅3+C2n⋅32=376

$$\Rightarrow 1-3n+\frac{n(n-1)}{2}\cdot 9=376$$⇒1−3n+2n(n−1)​⋅9=376

$$\Rightarrow 1-3n+\frac{{n}^{2}-n}{2}\cdot 9=376$$⇒1−3n+2n2−n​⋅9=376

$$\Rightarrow 2-6n+9{n}^{2}-9n=752$$⇒2−6n+9n2−9n=752

$$\Rightarrow 9{n}^{2}-15n-750=0$$⇒9n2−15n−750=0

$$\Rightarrow 3{n}^{2}-5n-250=0$$⇒3n2−5n−250=0

$$\Rightarrow n=\frac{5\pm \sqrt{25+3000}}{6}$$⇒n=65±25+3000​​

$$\Rightarrow n=\frac{5\pm 55}{6}$$⇒n=65±55​

$$\Rightarrow n=10{\text{ignoring negative sign}}$$⇒n=10ignoring negative sign

Now, ${T}_{r+1}={(-1)}^{r}Cr10{3}^{r}{x}^{10-3r}$Tr+1​=(−1)rCr103rx10−3r

So, for coefficient of ${x}^{4}$x4 we take

$$10-3r=4$$10−3r=4

$$\Rightarrow 3r=6$$⇒3r=6

$$\Rightarrow r=2$$⇒r=2

So, coefficient of ${x}^{4}$x4 is given by,

${T}_{2+1}={(-1)}^{2}\cdot C210{3}^{2}=45\times 9=405$T2+1​=(−1)2⋅C21032=45×9=405.