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7.Binomial Theorem
normal
If the coefficients of $x^7$ & $x^8$ in the expansion of ${\left[ {2\,\, + \,\,\frac{x}{3}} \right]^n}$ are equal , then the value of $n$ is :
A
$15$
B
$45$
C
$55$
D
$56$
Solution
$2^n {\left( {1\,\, + \,\,\frac{x}{6}} \right)^n} $
$\Rightarrow T_{r + 1} = 2^n . ^nC_r {\left( {\frac{x}{6}} \right)^r}$
$\Rightarrow 2^n . ^nC_7 . \frac{1}{{{6^7}}} = 2^n $. $^nC_8 .\frac{1}{{{6^8}}} $
$\Rightarrow 6 . ^nC_7 = ^nC_8 $
$\Rightarrow\,\, n – 7 = 48$
$ \Rightarrow n = 55$
Standard 11
Mathematics
