数列1/2*5,1/5*2,1/8*1,...,1/(3n-1)(3n+2),...求它的前n项和_数学解答

数列1/2*5,1/5*2,1/8*1,...,1/(3n-1)(3n+2),...求它的前n项和

人气：224 ℃　时间：2020-01-26 02:37:56

解答

1/(3n-1)(3n+2)=1/3*[1/(3n-1)-1/(3n+2)]
Σ1/(3n-1)(3n+2)=1/3*{[1/2-1/5]+[1/5-1/8]+...+[1/(3n-4)-1/(3n-1)]+[1/(3n-1)-1/(3n+2)]}
=1/3[1/2-1/(3n+2)]
=n/(6n+4)Σ什么意思Σ表示求和 Σ1/(3n-1)(3n+2) 代表 1/2*5+1/5*8*+1/8*11+...+1/(3n-1)(3n+2)的意思