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7.Binomial Theorem
hard
ધારોકે $\left(x-\frac{3}{x^2}\right)^n, x \neq 0 . n \in N$ ના વિસ્તરણમાં પ્રથમ ત્રણ પદોના સહગુણકોનો સરવાળો $376$ છે. તો $x^4$ નો સહગુણક $..........$ છે.
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
