Question

高二数学 求详细解

Answer 1

用裂项相消法，因为1/(n^2+2n)=1/n(n+2)=1/2(1/n-1/(n+2));
1/(1^2+2)=1/2*(1-1/3);1/(2^2+4)=1/2*(1/2-1/4).....
所以前n项和为1/2*(1-1/3)+1/2*(1/2-1/4)+1/2*(1/3-1/5)+...+1/2(1/(n-2)-1/(n))+1/2(1/(n-1)-1/(n+1))+1/2(1/n-1/(n+2))=1/2*(1+1/2-1/(n+1)-1/(n+2))=3/4--1/(2(n+1)）-1/(2(n+2))

Answer 2

数列的通项为an=1/(n²+2n);前n项和为Sn；
当n=1时，Sn=1/3;

当n≥2时；an=1/(n²+2n)= [1/n - 1/(n+2)]/2 ;
Sn= 1/2[1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7+..............+1/n - 1/(n+2)]
=1/2[1-1(n+1)-1/(n+2)]
=1/2
1/(n+1) 和1/(n+2) 在n趋于无穷时都为0

Answer 3

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Answer 4

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