6 t h term in expansion of ( 2 x 2 − 1 3 x 2 ) 10 is Options: A . &nbsp 4580 17 B . &nbsp - 896 27 C . &nbsp 5580 17 D . &nbsp None of these

Answer: Option B : B Applying T r + 1 = n C r x n − r d for ( x + a ) n Hence T 6 = 10 C 5 ( 2 x 2 ) 5 (- 1 3 x 2 ) 5 = - ( 10 ! 5 ! 5 ! ) (32) ( 1 243 ) = - 896 27

6th term in expansion of (2x2−13x2)10 is

Question

$6^{t h}$ term in expansion of $(2 x^{2} - \frac{1}{3 x^{2}})^{10}$ is

Options:

A .

\frac{4580}{17}

B . -

\frac{896}{27}

C .

\frac{5580}{17}

D . None of these

Answer: Option B
:
B

Applying $T_{r + 1}$ = $^{n} C_{r}$ $x^{n - r} d$ for $(x + a)^{n}$

Hence $T_{6}$ = $^{10} C_{5}$ $(2 x^{2})^{5}$ (- $\frac{1}{3 x^{2}})^{5}$

= - $(\frac{10!}{5! 5!})$ (32) $(\frac{1}{243})$ = - $\frac{896}{27}$

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