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EXERCISE 4.3 · Q36

Q.In the following expansion, find the indicated term: (4x5−52x)9\left(\dfrac{4x}{5}-\dfrac{5}{2x}\right)^9, 7th term.

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✓ Free question

Here a=4x5,b=−52x,n=9a=\dfrac{4x}{5},b=-\dfrac{5}{2x},n=9. For t7t_7, r=6r=6. t7=9C6(4x5)3(−52x)6=84⋅64x3125⋅1562564x6=84⋅15625125⋅x−3=84(125)x−3=10500x3t_7={}^9C_6\left(\dfrac{4x}{5}\right)^3\left(-\dfrac{5}{2x}\right)^6=84\cdot\dfrac{64x^3}{125}\cdot\dfrac{15625}{64x^6}=84\cdot\dfrac{15625}{125}\cdot x^{-3}=84(125)x^{-3}=\dfrac{10500}{x^3}.

✓Final answer

The 7th term is 10500x3\dfrac{10500}{x^3}.

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