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7.Binomial Theorem
medium
${\left( {2{x^2} - \frac{1}{{3{x^2}}}} \right)^{10}}$ ના વિસ્તરણ ${6^{th}}$ પદ મેળવો.
A
$\frac{{4580}}{{17}}$
B
$ - \frac{{896}}{{27}}$
C
$\frac{{5580}}{{17}}$
D
એકપણ નહીં.
Solution
(b) Applying ${T_{r + 1}} = {\,^n}{C_r}{x^{n – r}}{a^r}$ for ${(x + a)^n}$
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}}$
Standard 11
Mathematics
