已知递增的等比数列{an}的前三项之积为512,且这三项分别依次减去1、3、9后又成等差数列,求(1)数列{an}的通项公式an;(2)求证(1/a1)+(2/a2)+(3/a3)+...+(n/an)
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![已知递增的等比数列{an}的前三项之积为512,且这三项分别依次减去1、3、9后又成等差数列,求(1)数列{an}的通项公式an;(2)求证(1/a1)+(2/a2)+(3/a3)+...+(n/an)](/uploads/image/z/1160521-25-1.jpg?t=%E5%B7%B2%E7%9F%A5%E9%80%92%E5%A2%9E%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8D%E4%B8%89%E9%A1%B9%E4%B9%8B%E7%A7%AF%E4%B8%BA512%2C%E4%B8%94%E8%BF%99%E4%B8%89%E9%A1%B9%E5%88%86%E5%88%AB%E4%BE%9D%E6%AC%A1%E5%87%8F%E5%8E%BB1%E3%80%813%E3%80%819%E5%90%8E%E5%8F%88%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E6%B1%82%EF%BC%881%EF%BC%89%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8Fan%EF%BC%9B%EF%BC%882%EF%BC%89%E6%B1%82%E8%AF%81%281%2Fa1%29%2B%282%2Fa2%29%2B%283%2Fa3%29%2B...%2B%28n%2Fan%29)
已知递增的等比数列{an}的前三项之积为512,且这三项分别依次减去1、3、9后又成等差数列,求(1)数列{an}的通项公式an;(2)求证(1/a1)+(2/a2)+(3/a3)+...+(n/an)
已知递增的等比数列{an}的前三项之积为512,且这三项分别依次减去1、3、9后又成等差数列,求
(1)数列{an}的通项公式an;
(2)求证(1/a1)+(2/a2)+(3/a3)+...+(n/an)
已知递增的等比数列{an}的前三项之积为512,且这三项分别依次减去1、3、9后又成等差数列,求(1)数列{an}的通项公式an;(2)求证(1/a1)+(2/a2)+(3/a3)+...+(n/an)
(1)
由于前三项之积为512
所以:(a1)(a2)(a3) = (a2/q)(a2)(a2q) = (a2)³ = 512
因此:a(2)=8
且:a(1)-1,a(2)-3,a(3)-9成等差数列:
\x09[a(1)-1] + [a(3)-9] = 2[a(2)-3]
即:\x09a(2)/q - 1 + a(2)*q - 9 =2a(2) - 6
即:\x098/q +8q -10=10
即:\x092q²-5q +2=0
解出\x09q=2或q=1/2
而由于等比数列为递增的,所以q=2
因此a1=4,a2=8,a2=16
等比数列通项公式:
\x09a(n) = 2^(n+1)
(2)
令
S = (1/a1)+(2/a2)+(3/a3)+...+(n/an)
= 1/4 + 2/8 + ...+ (n-1)/2^n + n/2^(n+1)
则:
2S = 1/2 + 2/4 +3/8 ...+ n/2^n
所以
S = 2S - S
= 1/2 + 1/4 + 1/8 + ...+ 1/2^n - n/2^(n+1)
= (1/2)[1-1/2^n]/[1-(1/2)] - n/2^(n+1)
=1 - 1/2^n - n/2^(n+1)
=1 - (n+2)/2^(n+1)