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人教A版选修2-3第40页第8(3)题:已知(1+(?))~n的展开式中第9项,第10项、第11项的二项式系数成等差数列,求n.可将上题简化为:三个连续的二项式系数C_n~8,C_n~9,C_n~(10)成等差数列,求n.(n=14或23)让学生动手求出n,n=14或23,即C_(23)~8,C_(23)~9,C_(23)~(10)和C_(14)~8,C_(14)~9,C_(14)~(10)分别成等差数列.其中前者公差为正,后者公差为负.根据组合数性质C_n~m=C_n~(n-m),由前者可得C_(23)~(13),C_(23)~(14),C_(23)~(15)成等差数列,公差为负.由后者可得C_(14)~4,C_(14)~5,C_(14)68成等差数列,公差为正.为了研究方便,本课只
People teach A version of elective 2-3 p. 40, Section 8 (3): Known (1 + (?)) ~ N Expansion of the 9th, 10th, 11 binomial coefficients (N = 14 or 23), we can simplify the problem as follows: three consecutive binomial coefficients C_n ~ 8, C_n ~ 9, C_n ~ Let students learn n, n = 14 or 23, namely C_ (23) ~ 8, C_ (23) ~ 9, C_ (23) ~ (10) and C_ (14) ~ 8, C_ (14) ~ 9 , C_ (14) ~ (10) are respectively equal difference series, of which the former is positive and the latter is negative. According to the number of combinations C_n ~ m = C_n ~ (13), C_ (23) ~ (14) and C_ (23) ~ (15), and the tolerance is negative. From the latter, C14-14, C14-14, (14) 68 into the arithmetic sequence, the tolerance is positive. In order to facilitate the study, this lesson only