解:由于数列an是等差数列,所以:
(等差数列求和我喜欢用:(首项 + 末项)乘以项数/2,你用公式法也可以)
S3 = 3(a1 + a3)/2 , S5 = 5(a1 + a5)/2,都代入 S3 + S5 = 21,解出a5为:
a5 = [ 42 - (8a1 + 3a3) ] /5; (1)
另外,根据bn = 1/Sn, 所以b3 = 1/S3,于是
a3b3 = a3/S3 = a3/ [3(a1 + a3) /2 ] = (2/3) a3/(a1+a3) = 1/2,于是解出a3为:
a3 = 3a1,(2)
将这一表达式代入(1),得到:
a5 = [ 42 - 17a1] / 5. (3)
再利用等差数列通项公式,设公差为d,则有a3 = a1 + 2d, a5 = a1 + 4d,都代入到(2)式和(3)式中,就可以解出:
a1 = d = 1.
于是Sn = (a1 + an)n / 2 = [1 + (1 + (n-1)*1) ]n/2 = (n+1)/2,从而
bn = 1/Sn = 2/(n+1)。
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