Let the coefficient of X3 in (AX2 + 1x) & nbsp; 4 expansion be 32, then LiMn →∞ (a + A2 +...) +an)=（　　） A. 14B. 12C. 2D. 1

The general term of (AX2 + LX) 4 expansion is tr + 1 = A4 − rcr4x8 − 5r2. Let 8 − 52r = 3 get r = 2. The coefficient of X3 in the expansion is a2c24 = 32. The solution is a = 12  LiMn →∞ (a + A2 +...) +An) = LiMn →∞ 12 (1 − (12) n) 1 − 12 & nbsp; = 1, so D is selected