已知数列An=2^(n+1),求(1/A1)+(2/A2)+(3/A3)+......+(n/An)的和。

显示全部楼层 · 2011-7-18 17:17:16

An=2^(n+1),所以，设Bn=n/An=n/2^(n+1)则Sn=(1/A1)+(2/A2)+(3/A3)+......+(n/An)=B1+B2+B3+....+Bn知道，Sn=1/(2^2)+2/(2^3)+3/(2^4)+...+n/2^(n+1)---------------式1两边同乘以2得2Sn=1/(2^1)+2/(2^2)+3/(2^3)+...+n/(2^n)---------------式2所以，式2减式1得到Sn=1/(2^1)+1/(2^2)+1/(2^3)+...+1/(2^n)-n/2^(n+1)=1-(1+n/2)/(2^n)...

千问 · 2011-7-18 17:17:16

A=1/2^2+2/2^3+……+n/2^n+1
1/2A=1/2^3+2/ 2^4+……+n/2^n+2A-1/2A=1/2A=1/2^2+1/2^3+……+1/2^n+1-1/2^n+2
1/2^2+1/2^3+……+1/2^n+1为等比数列，用前n项和公式求A=2*（求完之后的式子-1/2^n+2）...

千问 · 2011-7-18 17:17:16

1/An是等比数列设所求为sn则2sn = 1/A0 + 2/A1 + ... + n/An-1求2sn - sn即可...