MCQ
${\left( {2{x^2} - \frac{1}{{3{x^2}}}} \right)^{10}}$ ના વિસ્તરણ ${6^{th}}$ પદ મેળવો.
- A$\frac{{4580}}{{17}}$
- ✓$ - \frac{{896}}{{27}}$
- C$\frac{{5580}}{{17}}$
- Dએકપણ નહીં.
Hence ${T_6} = {\,^{10}}{C_5}{(2{x^2})^5}{\left( { - \frac{1}{{3{x^2}}}} \right)^5}$
$ = - \frac{{10\,!}}{{5\,!\,5\,!}}32 \times \frac{1}{{243}} = - \frac{{896}}{{27}}$
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