Question

两道数学题，急！！！！！！！！！

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)

Answer 1

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

Answer 2

提示，1/n*(n+3)=(1/3)[3/(n*(n+3))]=(1/3)(1/n-1/(n+3))

Answer 3

1,
1/n*(n+3)=1/3(1/n-1/(n+3));
原式=1/3(1-1/4+1/2-1/5+1/3-1/6+1/4-1/7.....1/n-1/(n+3))=1/3(1+1/2+1/3-1/(n+1)-1/(n+2)+1/(n+3))=
2,
1/n(n+2)=1/2(1/n-1/(n+2));
原式=1/2(1+1/2-1/(n+1)-1/(n+2))=结果
总结：1/n(n+m)=1/m(1/n-1/(n+m))