7.Binomial Theorem
easy

यदि ${\left( {{x^2} + \frac{1}{x}} \right)^n}$ के विस्तार में मध्य पद $924{x^6}$ हो, तो $n = $

A

$10$

B

$12$

C

$14$

D

इनमें से कोई नहीं

Solution

$n$ सम है अत: $\left( {\frac{n}{2} + 1} \right)\,oka$ पद मध्य पद होगा

अत: $^n{C_{n/2}}{({x^2})^{n/2}}{\left( {\frac{1}{x}} \right)^{n/2}} = 924{x^6}$

$ \Rightarrow $ ${x^{n/2}} = {x^6} \Rightarrow n = 12$.

Standard 11
Mathematics

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