人教版高中数学第三册选修2,100页 求和1/1*4+1/4*7+1/7*10+...+1/(3K-2)(3K+1)+1

1/(3K-2)(3K+1)=1/3*3/(3K-2)(3K+1)=1/3*[(3K+1)-(3K-2)]/(3K-2)(3K+1)=1/3*[1/(3K-2)-1/(3K+1)]
所以
1/1*4+1/4*7+1/7*10+...+1/(3K-2)(3K+1)+1/[3(K+1)-2][3(K+1)+1]
=1/3*(1-1/4)+1/3*(1/4-1/7)+1/3*(1/7-1/10)+...+1/3*[1/(3K-2)-1/(3K+1)]+1/3*[1/(3K+1)-1/(3K+4)]
=1/3*[1-1/4+1/4-1/7+1/7-1/10+.+1/(3K-2)-1/(3K+1)+1/(3K+1)-1/(3K+4)]
=1/3*[1-1/(3K+4)]
=1/3*(3K+3)/(3K+4)
=(K+1)/(3K+4)
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