#r# th term of the given series

#t_r=1/((2,5,8...r"th term")(5,8,11...r"th term"))#

#=>t_r=1/((2+(r-1)xx3)(5+(r-1)xx3))#

#=>t_r=1/((3r-1)(3r+2))#

#=>t_r=1/3xx3/((3r-1)(3r+2))#

#=>t_r=1/3((3r+2)-(3r-1))/((3r-1)(3r+2))#

#=>t_r=1/3(1/(3r-1)-1/(3r+2))#

Putting #r=1,2,3....".upto " n# and adding we get

For #r=1," "" "# #t_1=1/3(1/2-cancel(1/5))#

For #r=2," "" "# #t_2=1/3(cancel(1/5)-cancel(1/8))#

For #r=3," "" "# #t_3=1/3(cancel(1/8)-cancel(1/11))#

#---------#

#---------#

#---------#

For #r=n-1," "# #t_(n-1)=1/3(cancel(1/(3n-4))-cancel(1/(3n-1)))#

For #r=n," "" "##t_n=1/3(cancel(1/(3n-1))-1/(3n+2))#

#"Sum upto n terms "S_n#

#=1/3(1/2-1/(3n+2))=(3n+2-2)/(3*2*(3n+2))=n/(2*(3n+2))#