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7.Binomial Theorem
easy
જો ${\left( {{x^2} + \frac{1}{x}} \right)^n}$ ના વિસ્તરણમાં મધ્યમપદ $924{x^6}$ હોય તો $n = $
A
$10$
B
$12$
C
$14$
D
એકપણ નહીં.
Solution
(b) Since $n$ is even therefore ${\left( {\frac{n}{2} + 1} \right)^{th}}$ term is middle term,
hence $^n{C_{n/2}}{({x^2})^{n/2}}{\left( {\frac{1}{x}} \right)^{n/2}} = 924{x^6}$
$ \Rightarrow $ ${x^{n/2}} = {x^6} \Rightarrow n = 12$.
Standard 11
Mathematics
