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7.Binomial Theorem
hard
જો $(1+x)^{p}(1-x)^{q}, p, q \leq 15$ ના વિસ્તરણમાં $x$ અને $x^{2}$ ના સહગુણકો અનુક્રમે $-3$ અને $-5$ હોય તો $x ^{3}$ નો સહગુણક $............$ થાય.
A
$22$
B
$23$
C
$52$
D
$53$
(JEE MAIN-2022)
Solution
Since coefficient of $x$ is $-3$
${ }^{p} C _{1}-{ }^{9} C _{1}=-3$
$p – q =-3$
$\text { Comparing coefficients of } x ^{2}$
${ }^{9} C _{1}{ }^{9} C _{1}+{ }^{ p } C _{2}+{ }^{9} C _{2}=-5$
$- pq +\frac{ p ( p -1)}{2}+\frac{ q ( q -1)}{2}=-5$
Solving $(1)$ and $(2)$
$p=8, q=11$
Coefficient of $x ^{3}$ is
$-{ }^{4} C_{3}+{ }^{P} C_{3}+{ }^{P} C_{1}^{9} C_{2}-{ }^{P} C_{2}^{9} C_{1}$
$=-{ }^{11} C_{3}+{ }^{8} C_{3}+{ }^{8} C_{1}^{11} C_{2}-{ }^{8} C_{2}^{11} C_{1}$
$=23$
Standard 11
Mathematics
