đặt
`A=1/9 +1/a^2+1/(a+3)^2`
`⇔A=1/9 +1/a^2+1/(a+3)^2+(2(a+3)-2a)/(a(a+3))-6/(a(a+3))`
`⇔A=1/9 +1/a^2+1/(a+3)^2+2/a-2/(a+3)-6/(a(a+3))`
`⇔A=(1/3+1/a-1/(a+3)^2)^2`
`\sqrt(1/9 +1/a^2+1/(a+3)^2)=\sqrt((1/3+1/a-1/(a+3))^2)=1/3+1/a-1/(a+3)`
áp dụng vào
`⇒S=1/3+1/1-1/4+1/3+1/4-1/9+...+1/3 +1/(46)-1/(49)`
`⇔S=(46)/3-1/(49)`
`⇔S=(2251)/(147)`