Đáp án:
Giải thích các bước giải:
`A=1/51+1/52+1/53+...+1/150`
`=>A=(1/51+1/52+...+1/75)+(1/76+1/77+...+1/150)`
`=>1/51>1/75;1/52>1/75;...;1/74>1/75`
`=>1/51+1/52+...+1/75>1/75+1/75+...+1/75` (Có `25` số hạng)
`=>1/51+1/52+...+1/75>1/3(1)`
`=>1/76>1/150;1/77>1/150;...;1/149>1/150`
`=>1/76+1/77+...+1/150>1/150+1/150+...+1/150` (Có `75` số hạng)
`=>1/76+1/77+...+1/150>1/2(2)`
Từ `(1)` và `(2)` ta có:
`=>1/51+1/52+1/53+...+1/150>1/2+1/3`
`=>1/51+1/52+1/53+...+1/150>5/6`
Vậy `A>5/6`.
`A=1/2^2+1/3^2+1/4^2+...+1/100^2`
`=>A<1/1.2+1/2.3+1/3.4+...+1/99.100`
`=>A<1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100`
`=>A<1-1/100<1`
`=>A<1`
Vậy `A<1`.
`A=1/2^2+1/3^2+1/4^2+....+1/100^2`
`=>A>1/2.3+1/3.4+1/4.5+...+1/100.101`
`=>A>1/2-1/3+1/3-1/4+1/4-1/5+...+1/100-1/101`
`=>A>1/2-1/101`
`=>A>1/2`
Vậy `A>1/2`.
