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Question 76 of 95

Q.Find the coefficient of x3x^3 in the expansion of (2−3x)7(2-3x)^7.

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2019Subjective· 3mImportance★★★★★
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The general term Tr+1=(7r)27−r(−3)rxrT_{r+1}=\binom{7}{r}2^{7-r}(-3)^r x^r gives, at r=3r=3, a coefficient of (73)⋅24⋅(−3)3=−15120\binom{7}{3}\cdot2^4\cdot(-3)^3 = -15120.

Expand (2−3x)7(2-3x)^7 using the binomial theorem: general term Tr+1=(7r)(2)7−r(−3x)r=(7r)27−r(−3)rxrT_{r+1} = \binom{7}{r}(2)^{7-r}(-3x)^r = \binom{7}{r}2^{7-r}(-3)^r x^r.

For the coefficient of x3x^3, set r=3r=3:

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