17.18
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17.1817.18 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+
17.18
17.18
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,...