Answer The Following Question ?

The general term can be written:

`a_n = 1/((3n-2)(3n+1)) = 1/3(1/(3n-2)-1/(3n+1))`

So:

`sum_(n=1)^N a_n = sum_(n=1)^N 1/3(1/(3n-2)-1/(3n+1))`

`color(white)(sum_(n=1)^N a_n) = 1/3(sum_(n=1)^N 1/(3n-2)-sum_(n=1)^N 1/(3n+1))`

`color(white)(sum_(n=1)^N a_n) = 1/3(sum_(n=1)^N 1/(3n-2)-sum_(n=2)^(N+1) 1/(3n-2))`

`color(white)(sum_(n=1)^N a_n) = 1/3(1+color(red)(cancel(color(black)(sum_(n=2)^N 1/(3n-2))))-color(red)(cancel(color(black)(sum_(n=2)^N 1/(3n-2)))) - 1/(3N+1))`

`color(white)(sum_(n=1)^N a_n) = 1/3(1-1/(3N+1))`

`color(white)(sum_(n=1)^N a_n) = N/(3N+1)`

Let ,

`S=1/(1*4)+1/(4*7)+1/(7*10)+...+1/((3n-2)(3n+1))`

`:.t_n=1/((3n-2)(3n+1))=1/3[1/(3n-2)-1/(3n+1)]`

`:.S_n=1/3sum_(r=1)^n [1/(3r-2)-1/(3r+1)]`

`:.S_n=1/3[(1/1-cancel(1/4))color(white)(........................)lArrcolor(red)( r=1`

`color(white)(..............)+(cancel(1/4)-cancel(1/7))color(white)(........................)lArr color(red)(r=2`

`color(white)(..............)+(cancel(1/7)-cancel(1/10))color(white)(........................)lArrcolor(red)( r=3`

`color(white)(..............)+(cancel(1/10)-cancel(1/13))color(white)(........................)lArrcolor(red)( r=4`

`color(white)(..............)+cancel(..).................`

`color(white)(..............)+...................`

`color(white)(..............)+.............cancel(......`

`color(white)(..............)+(cancel(1/(3n-5))-cancel(1/(3n-2)))color(white)(..............)lArrcolor(red)( r=n-1`

`color(white)(..............)+(cancel(1/(3n-2))-1/(3n+1))]color(white)(...............)lArrcolor(red)( r=n`

`:.S_n=1/3[1/1-1/(3n+1)]`

`:.S_n=1/3[(3n+1-1)/(3n+1)]=1/3[(3n)/(3n+1)]`

`:.S_n=n/(3n+1)`