1/n(n+3)的通项和详细步骤

显示全部楼层 · 2013-3-5 14:30:48

Sn=1/(1*4)+1/(2*5)+1/(3*6)+1/(4*7)+...+1/(n-3)n+1/(n-2)(n+1)+1/(n-1)(n+2)+1/n(n+3)=1/3*[3/(1*4)+3/(2*5)+3/(3*6)+3/(4*7)+...+3/(n-3)n+3/(n-2)(n+1)+3/(n-1)(n+2)+3/n(n+3)]=1/3*[1-1/4+1/2-1/5+1/3-1/6+1/4-1/7...+1/(n-3)-1/n+1/(n-2)-1/(n+1)+1/(n-1)-1/(n+2)+1/n-1/(n+3)]=1/3*[1+1/2+1/3-1/(n+1)-1/(n+2)-1/(n+3)]=1/3*[11/6-1/(n+1)-...

千问 · 2013-3-5 14:30:48

1/n(n+3)=[1/n-1/(n+3)]/3其和为1-1/4+1/2-1/5+1/3-1/6+1/4-1/7+...+1/(n-3)-1/n+1/(n-2)-1/(n+1)+1/(n-1)-1/(n+2)+1/n-1/(n+3)从第四组1/4-1/7开始，能被消去，结果为[1+1/2+1/3-1/(n+1)-1/(n+2)-1/(n+3)...

千问 · 2013-3-5 14:30:48

详细步骤不好在上面写，给你提供个思路把上面1/n(n+3)拆分成两个数字相差即分解成三分之一倍的n分之一减去n加三分之一，，然后分别相加，，，很简单的...

1/n(n+3)的通项和 详细步骤

1/n(n+3)的通项和详细步骤