MCQ
${\left( {\frac{{3{x^2}}}{2} - \frac{1}{{3x}}} \right)^9}$ ના વિસ્તરણમાં અચળપદ મેળવો.
- ✓$^9{C_3}.\frac{1}{{{6^3}}}$
- B$^9{C_3}{\left( {\frac{3}{2}} \right)^3}$
- C$^9{C_3}$
- Dએકપણ નહીં.
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${T_{6 + 1}} = {\,^9}{C_6}{\left( {\frac{3}{2}} \right)^{9 - 6}}{\left( { - \frac{1}{3}} \right)^6} = {\,^9}{C_3}.\frac{1}{{{6^3}}}$
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(જ્યાં $[.]$ એ મહતમ પૂર્ણાક વિધેય છે)
વિધાન $1$ : $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ નું વિચરણ $4{\sigma ^2}$ છે.
વિધાન $2$: $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ નો સમાંતર મધ્યક $4\bar x$ છે.