If the sum of the coefficients in the expansion of (3x − 13x2) n is 128, then the coefficient of 1x3 in the expansion is () A. 7B. -7C. 21D. -21

Let x = 1 get the sum of the coefficients of the expansion 2n, let 2n = 128, the solution is n = 7. Let (3x − 13x2) n = (3x − 13x2) 7, the general term of the expansion is tr + 1 = (− 1) r37 − rcr7x7 − 5r3, let 7 − 5r3 = − 3, the solution is r = 6. Therefore, the coefficient of 1x3 in the expansion is 3c76 = 21

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