(30 18:34:42) It is known that the first term A1 = 1, tolerance D is greater than 0, and the second term, the fifth term and the fourteenth term of an arithmetic sequence are the second term, the third term and the fourth term respectively (1) (2) let BN = 1 / N (an + 3) & # 160; (n belongs to n *), Sn = B1 + B2 + B3 + ·· BN to find SN  

(30 18:34:42) It is known that the first term A1 = 1, tolerance D is greater than 0, and the second term, the fifth term and the fourteenth term of an arithmetic sequence are the second term, the third term and the fourth term respectively (1) (2) let BN = 1 / N (an + 3) & # 160; (n belongs to n *), Sn = B1 + B2 + B3 + ·· BN to find SN  

According to the formula of arithmetic sequence, an = 1 + (n-1) d. so A2 = 1 + D, A5 = 1 + 4D, A14 = 1 + 13D (A5) ^ 2 = A2 * A14, that is, (1 + 4D) ^ 2 = (1 + D) (1 + 13D), d > 0, so d = 2An = 1 + 2 (n-1) = 2n-12. BN = 1 / [n (2n-1 + 3)] = 1 / [2n (n + 1)] = (1 / 2) [(1 / N) - (1 / N + 1)] Sn = (1