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7.Binomial Theorem
easy
${\left( {2 + \frac{x}{3}} \right)^n}$ ના વિસ્તરણમાં ${x^7}$ અને ${x^8}$ ના સહગુણક સમાન હોય તો . . . .
A
$56$
B
$55$
C
$45$
D
$15$
Solution
(b) ${x^7}$,${x^8}$ will occur in ${T_8}$ and${T_9}$.
Coefficients of ${T_8}$ and ${T_9}$ are equal.
$\therefore {\,^n}{C_7}{2^{n – 7}}{\left( {\frac{1}{3}} \right)^7} = {\,^n}{C_8}{2^{n – 8}}{\left( {\frac{1}{3}} \right)^8}\,\, $
$\Rightarrow n = 55$.
Standard 11
Mathematics
