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7.Binomial Theorem
easy
${\left( {\frac{{3{x^2}}}{2} - \frac{1}{{3x}}} \right)^9}$ ના વિસ્તરણમાં અચળપદ મેળવો.
A
$^9{C_3}.\frac{1}{{{6^3}}}$
B
$^9{C_3}{\left( {\frac{3}{2}} \right)^3}$
C
$^9{C_3}$
D
એકપણ નહીં.
Solution
(a) In the expansion of ${\left( {\frac{{3{x^2}}}{2} + \frac{1}{{3x}}} \right)^9}$,
the general term is ${T_{r + 1}} = {\,^9}{C_r}.{\left( {\frac{{3{x^2}}}{2}} \right)^{9 – r}}{\left( { – \frac{1}{{3x}}} \right)^r}$
$ = {\,^9}{C_r}{\left( {\frac{3}{2}} \right)^{9 – r}}{\left( { – \frac{1}{3}} \right)^r}{x^{18 – 3r}}$
For the term independent of $x$, $18 -3r = 0$ ==> $ r = 6$
This gives the independent term${T_{6 + 1}} = {\,^9}{C_6}{\left( {\frac{3}{2}} \right)^{9 – 6}}{\left( { – \frac{1}{3}} \right)^6} = {\,^9}{C_3}.\frac{1}{{{6^3}}}$
Standard 11
Mathematics
