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MISCELLANEOUS EXERCISE 4 (II) · Q102

Q.Find third term in the expansion of (9x2−y36)4\left(9x^2-\dfrac{y^3}{6}\right)^4.

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Here a=9x2,b=−y36,n=4a=9x^2,b=-\dfrac{y^3}6,n=4. For t3t_3, r=2r=2. $t_3={}^4C_2(9x^2)^2\left(-\dfrac{y^3}6\right)^2=6(81x^4)\left(\dfrac{y^6}{36}\right)=6\times\dfrac{81}{36}x^4y^ …

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