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7.Binomial Theorem
medium
ધારો કે $(1+x)^{2 n -1}$ ના દ્વિપદ્દી વિસ્તરણમાં $30$ માં અને $12$ માં પદોના સહગુણકો અનુક્રમ $A$ અને $B$ છે. ને $2 A=5 B$ હોય, તો $n =$ _______
A$22$
B$21$
C$20$
D$19$
(JEE MAIN-2025)
Solution
$A={ }^{2 n-1} C_{29} B={ }^{2 n-1} C_{11}$
$2{ }^{2 n-1} C_{29}=5{ }^{2 n-1} C_{11}$
$2 \frac{(2 n-1)!}{29!(2 n-30)!}=5 \frac{(2 n-1)!}{(2 n-12)!11!}$
$\frac{1}{29 \ldots 12 \cdot 5}=\frac{1}{(2 n-12)(2 n-13) \ldots(2 n-29) 2}$
$\frac{1}{30 \cdot 29 \ldots 12}=\frac{1}{(2 n-12)(2 n-13) \ldots(2 n-29) 12}$
$2 n-12=30$
$n=21$
$2{ }^{2 n-1} C_{29}=5{ }^{2 n-1} C_{11}$
$2 \frac{(2 n-1)!}{29!(2 n-30)!}=5 \frac{(2 n-1)!}{(2 n-12)!11!}$
$\frac{1}{29 \ldots 12 \cdot 5}=\frac{1}{(2 n-12)(2 n-13) \ldots(2 n-29) 2}$
$\frac{1}{30 \cdot 29 \ldots 12}=\frac{1}{(2 n-12)(2 n-13) \ldots(2 n-29) 12}$
$2 n-12=30$
$n=21$
Standard 11
Mathematics
