若Sn是数列an的前n项和,a1=3,2Sn=na(n+1)-n(n+1)(n+2)
问题描述:
若Sn是数列an的前n项和,a1=3,2Sn=na(n+1)-n(n+1)(n+2)
(1)求数列an的通项公式
(2)证明:对一切正整数n,有12/a1+12/a2+.+12/an<7
答
解决方案:⑴阶数n = 1,2 * S1-2 * A1 = A + B ==> A + B = 0 阶数N = 2,2 * S2-3 * A2 = 2A + B ==> 2A + B = -1 所以A = -1,B = 1 ⑵证明:⑴,然后2SN-(N +1)的= 1-N..①2 * S(N +1) - (N +2)* A(N +1)=-N...