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例1已知{a_n}为等差数列,以S_n表示{a_n}的前n项和,且a_1>0,S_9=S_(31),则使S_n取得最大值的n是A.21 B.20 C.19 D.18简单解法1(化基本量法)∵S_9=S_(31),∴9a_1+36d=31a_1+465d.∴a_1=-(39)/2d>0.∵S_n=na_1+(n(n-n))/2 d=d/2(n~2-40n)=d/2[(n-20)~2-400],又d<0,n∈N~*,∴当n=20时,S_n取得最大值.选B.简单解法2(数形结合法)∵S_n=An~2+Bn(A≠0),∴(n,S_n)为二次函数y=Ax~2+Bx(A≠0)的图像上一群孤立的点.∵a_1>0,S_9=S_(31),∴此二次函数开口向
Example 1 It is known that {a_n} is an arithmetic progression, and S_n represents the first n terms of {a_n}, and a_1> 0 and S_9 = S_ (31). 20 C.19 D.18 Simple Solution 1 (Basicization Method) ∵S_9 = S_ (31), ∴9a_1 + 36d = 31a_1 + 465d.∴a_1 = - (39) / 2d> 0.∵S_n = na_1 + ( n nN / 2 d = d / 2 n ~ 2-40n = d / 2 [(n-20) ~ 2-400] 20, the maximum value of S_n is chosen. B. Simple Solution 2 (Combination Method) ∵S_n = An ~ 2 + Bn (A ≠ 0), ∴ (n, S_n) is quadratic function y = Ax ~ 2 + A group of isolated points on the image of Bx (A ≠ 0) .∵a_1> 0, S_9 = S_ (31), ∴ This quadratic function opens to