数学的数列求和 dn=2^(2n-1)+1,求证1/(d2-d1)+1/(d3-d2)+...+1/(d(n+1)-dn)
问题描述:
数学的数列求和 dn=2^(2n-1)+1,求证1/(d2-d1)+1/(d3-d2)+...+1/(d(n+1)-dn)
数学人气:986 ℃时间:2019-10-25 10:15:32
优质解答
由通项可得1/(d2-d1)+1/(d3-d2)+...+1/(d(n+1)-dn)=1/(2^3-2^1)+1/(2^5-2^3)+1/(2^7-2^5)+.+1/[2^(2n+1)-2^(2n-1)]=1/3*[1/2+1/2^3+1/2^5+.+1/2^(2n-1)]=1/3*1/2*[1-(1/2)^(2n)]/(1-1/4)=2/9*[1-(1/2)^(2n)]
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答
由通项可得1/(d2-d1)+1/(d3-d2)+...+1/(d(n+1)-dn)=1/(2^3-2^1)+1/(2^5-2^3)+1/(2^7-2^5)+.+1/[2^(2n+1)-2^(2n-1)]=1/3*[1/2+1/2^3+1/2^5+.+1/2^(2n-1)]=1/3*1/2*[1-(1/2)^(2n)]/(1-1/4)=2/9*[1-(1/2)^(2n)]