7.Binomial Theorem
hard

माना $\left(\mathrm{x}-\frac{3}{\mathrm{x}^2}\right)^{\mathrm{n}}, \mathrm{x} \neq 0, \mathrm{n} \in \mathrm{N}$, के प्रसार में प्रथम तीन पदों के गुणांको का योग 376 है। तो $\mathrm{x}^4$ का गुणांक ___________ है।

A

$404$

B

$403$

C

$402$

D

$405$

(JEE MAIN-2023)

Solution

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

Sum of coefficients of first three terms

${ }^n C_0-{ }^n C_1 \cdot 3+{ }^n C_2 3^2=376$

$\Rightarrow 3 n^2-5 n-250=0$

$\Rightarrow(n-10)(3 n+25)=0$

$\Rightarrow n =10$

Now general term ${ }^{10} C _{ r } x ^{10- r }\left(\frac{-3}{ x ^2}\right)^{ r }$

$={ }^{10} C _{ r } x ^{10- r }(-3)^{ r } \cdot x ^{-2 r }$

$={ }^{10} C _{ r }(-3)^{ r } \cdot x ^{10-3 r }$

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

$\Rightarrow r =2$

${ }^{10} C _2(-3)^2=405$

Standard 11
Mathematics

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