Đáp án + giải thích các bước giải:
Đặt `A=1^2+2^2+3^2...+n^2`
Ta có: `n^2=(n-1)n+n`
`->A=(1-1).1+1+(2-1).2+2+(3-1).3+3...+(n-1)n+n`
`=1+1.2+2+2.3+3+...+(n-1).n+n`
`=1+2+...+n+1.2+2.3+3.4+...+(n-1).n`
`=((n+1)n)/2+1.2+2.3+3.4+...+(n-1).n`
Đặt `B=1.2+2.3+3.4+..+(n-1).n`
`->3B=1.2.3+2.3.3+3.4.3+...+(n-1).n.3`
`=1.2.3+2.3.(4-1)+3.4.(5-2)+...+(n-1).n.[n+1-(n-2)]`
`=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+(n-1).n.(n+1)-(n-1)(n-2)n`
`=(n-1)n(n+1)`
`->B=((n-1)n(n+1))/3`
`->A=((n-1)n(n+1))/3+((n+1)n)/2`
`=(2(n-1)n(n+1)+3n(n+1))/6`
`=(n(n+1)(2n-2+3))/6`
`=(n(n+1)(2n+1))/6`
`->lim (1^2+2^2+3^2...+n^2)/(n(n+1)(n+2))=lim((n(n+1)(2n+1))/6)/(n(n+1)(n+2))=lim(2n+1)/(6n+12)=lim(n(2+1/n))/(n(6+12/n))=2/6=1/3`
