1/1*4+1/2*5+1/3*6+……+1/n*(n+3) 证明1/1*3+1/2*4+1/3*5+……+1/n(n+2)=3/4-(2n+3)/2(n+1)(n+2)
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1/n(n+2)= 1/2*[1/n -1/(n+2)]
1/1*3+1/2*4+1/3*5+……+1/n(n+2)
=1/2*[ 1/1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+...+1/(n-2)-1/n+1/(n-1)-1/(n+1)+1/(n)-1/(n+2) ]
=1/2*{ [ 1/1+1/2+1/3+...+1/(n-1)+1/n]-[1/3+1/4+1/5+...+1/n+1/(n+1)+1/(n+2) }
=1/2*{ [ 1/1+1/2]-[1/(n+1)+1/(n+2) }
=3/4-(2n+3)/2(n+1)(n+2)
1/n(n+3)= 1/3*[1/n -1/(n+3)]
1/1*4+1/2*5+1/3*6+……+1/n*(n+3) 【同样处理】
=1/3*{ [ 1/1+1/2+1/3+...+1/(n-1)+1/n]-[1/4+1/5+...+1/n+1/(n+1)+1/(n+2)+1/（n+3) }
=1/3*{ [ 1/1+1/2+1/3]-[1/(n+1)+1/(n+2)+1/（n+3) }
= 11/18-[1/(n+1)+1/(n+2)+1/（n+3)]/3
1/1*3+1/2*4+1/3*5+……+1/n(n+2)
=1/2*[ 1/1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+...+1/(n-2)-1/n+1/(n-1)-1/(n+1)+1/(n)-1/(n+2) ]
=1/2*{ [ 1/1+1/2+1/3+...+1/(n-1)+1/n]-[1/3+1/4+1/5+...+1/n+1/(n+1)+1/(n+2) }
=1/2*{ [ 1/1+1/2]-[1/(n+1)+1/(n+2) }
=3/4-(2n+3)/2(n+1)(n+2)
1/n(n+3)= 1/3*[1/n -1/(n+3)]
1/1*4+1/2*5+1/3*6+……+1/n*(n+3) 【同样处理】
=1/3*{ [ 1/1+1/2+1/3+...+1/(n-1)+1/n]-[1/4+1/5+...+1/n+1/(n+1)+1/(n+2)+1/（n+3) }
=1/3*{ [ 1/1+1/2+1/3]-[1/(n+1)+1/(n+2)+1/（n+3) }
= 11/18-[1/(n+1)+1/(n+2)+1/（n+3)]/3
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