作业帮17.18
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/22 13:11:06
17.18
因为a(n) = a + (n-1)d,n = 1,2,...S(n) = na + n(n-1)d/2.b(n) = 1/S(n) = 1/[na + n(n-1)d/2] = 2/[2na + n(n-1)d],1/2 = a(3)b(3) = [a+2d]*2/[6a+6d] = (a+2d)/(3a+3d),3a+3d = 2a + 4d,a = d.b(n) = 2/[2na + n(n-1)d] = 2/[n(n+1)d].S(n) = na + n(n-1)d/2 = nd[2 + n-1]/2 = n(n+1)d/2.21 = S(3)+S(5) = 3*4d/2 + 5*6d/2 = 6d + 15d = 21d,d = 1,b(n) = 2/[n(n+1)d] = 2/[n(n+1)],n = 1,2,... 再答： 裂项求和:当公差为零时,显然成立.
当公差不为零时,因为1/anan+1=1/d*[1/an-1/an+1]
1/a1a2+1/a3a4+…+1/anan+1=1/d*[1/a1--1/an+1]=1/d*[(an+1-an)/a1an+1]
=1/d*[nd/a1an+1]=n/a1an+1
当公差不为零时,因为1/anan+1=1/d*[1/an-1/an+1]
1/a1a2+1/a3a4+…+1/anan+1=1/d*[1/a1--1/an+1]=1/d*[(an+1-an)/a1an+1]
=1/d*[nd/a1an+1]=n/a1an+1