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EXERCISE 4.3 · Q34

Q.In the following expansion, find the indicated term: (2x2+32x)8\left(2x^2+\dfrac{3}{2x}\right)^8, 3rd term.

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✓ Free question

Here a=2x2,b=32x,n=8a=2x^2,b=\dfrac{3}{2x},n=8. For t3t_3, r=2r=2. t3=8C2(2x2)6(32x)2=28(64x12)(94x2)=28×64×94x10=28(144)x10=4032x10t_3={}^8C_2(2x^2)^6\left(\dfrac{3}{2x}\right)^2=28(64x^{12})\left(\dfrac{9}{4x^2}\right)=28\times\dfrac{64\times9}{4}x^{10}=28(144)x^{10}=4032x^{10}.

✓Final answer

The 3rd term is 4032x104032x^{10}.

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