Let, #t_n# denote the #n^(th)# term of the series,
# s_n=1/(1*4)+1/(4*7)+1/(7*10)+... "to n terms."#
Observe that, the First Factors of the Dr. of #t_n# are
#1,4,7,... ,# which form an A.P., with the first term #a=1,# and the
common difference,
#d=3; :. n^(th) term=a+(n-1)d=1+3(n-1)=3n-2.#
Similarly, for the Second Factors, #n^(th) term=3n+1.#
#:. t_n=1/{(3n-2)(3n+1)}.#
#:. t_n=1/3{3/{(3n-2)(3n+1)}},#
#=1/3{{(3n+1)-(3n-2)}/{(3n-2)(3n+1)}},#
#=1/3{(3n+1)/{(3n-2)(3n+1)} -(3n-2)/{(3n-2)(3n+1)}},#
#=1/3{1/(3n-2)-1/(3n+1)},#
#:. s_n=1/3sum_1^n{1/(3n-2)-1/(3n+1)},#
#=1/3{(1/1-cancel(1/4))+(cancel(1/4)-cancel(1/7))+(cancel(1/7)-cancel(1/10))+...+(cancel(1/(3n-5))-cancel(1/(3n-2)))+(cancel(1/(3n-2))-1/(3n+1))},#
#=1/3{1/1-1/(3n+1)},#
#=1/3{{(3n+1)-1}/(3n+1)},#
# rArr s_n=n/(3n+1),# is the desired sum.
Enjoy Maths.!
