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MISCELLANEOUS EXERCISE 4 (II) · Q106

Q.Find the middle term (s) in the expansion of (x2+2y2)7(x^2+2y^2)^7.

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Here a=x2,b=2y2,n=7a=x^2,b=2y^2,n=7. Middle terms at r=3r=3 (t4t_4) and r=4r=4 (t5t_5). t4=7C3(x2)4(2y2)3=35x8(8y6)=280x8y6t_4={}^7C_3(x^2)^4(2y^2)^3=35x^8(8y^6)=280x^8y^6. $t_5={}^7C_4(x^2)^3(2y^2)^4=35x^6(16y^8 …

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