It is evident that sum of the series tends to #oo#. However, let us find sum up to #n# terms.
#n^(th)# term of the series #1*3*5+3*5*7+5*7*9+....# is
#(2n-1)(2n+1)(2n+3)# and we have to find #sum_1^n((2n-1)(2n+1)(2n+3))#
As #(2n-1)(2n+1)(2n+3)#
= #(2n+3)(4n^2-1)#
= #8n^3+12n^2-2n-3#
Hence #sum_1^n(8n^3+12n^2-2n-3)#
= #8sum_1^n n^3+12sum_1^n n^2-2sum_1^n n-3n#
= #8(n^2(n+1)^2)/4+12(n(n+1)(2n+1))/6-2(n(n+1))/2-3n#
= #2n^2(n+1)^2+2n(n+1)(2n+1)-n(n+1)-3n#
= #2n^2(n^2+2n+1)+(2n^2+2n)(2n+1)-n^2-n-3n#
= #2n^4+4n^3+2n^2+4n^3+2n^2+4n^2+2n-n^2-n-3n#
= #2n^4+8n^3+7n^2-2n#
It is apparent that as #n->oo# sum of the series tends to #oo#.
