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MISCELLANEOUS EXERCISE 4 (II) · Q104

Q.Find the middle term (s) in the expansion of (2a3−32a)6\left(\dfrac{2a}{3}-\dfrac{3}{2a}\right)^6.

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Here a=2a3,b=−32a,n=6a=\dfrac{2a}3,b=-\dfrac3{2a},n=6. Middle term at r=3r=3 (t4t_4). $t_4={}^6C_3\left(\dfrac{2a}3\right)^3\left(-\dfrac3{2a}\right)^3=20\cdot\dfrac{8a^3}{27}\cdot\left(-\dfr …

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