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7.Binomial Theorem
easy
${\left( {2x + \frac{1}{{3x}}} \right)^6}$ ના વિસ્તરણમાં અચળપદ મેળવો.
A
$\frac{{160}}{9}$
B
$\frac{{80}}{9}$
C
$\frac{{160}}{{27}}$
D
$\frac{{80}}{3}$
Solution
(c) $1(6 – r) + ( – 1)r = 0 \Rightarrow r = 3$,
therefore fourth term will be independent of $x$ i.e.
$^6{C_3}{(2x)^3}{\left( {\frac{1}{{3x}}} \right)^3} = 20 \times 8 \times \frac{1}{{27}} = \frac{{160}}{{27}}$
Standard 11
Mathematics
