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*2^2+2*4*3^2+3*5*4^2+....# is

#n(n+2)(n+1)^2# and we have to find #sum_1^n(n(n+2)(n+1)^2)#

As #n(n+2)(n+1)^2#

= #(n^2+2n)(n^2+2n+1)#

= #n^4+2n^3+n^2+2n^3+4n^2+2n#

= #n^4+3n^3+5n^2+2n#

Hence #sum_1^n(n^4+3n^3+5n^2+2n)#

= #sum_1^n n^4+3sum_1^n n^3+5sum_1^n n^2+2sum_1^n n#

= #(n(n+1)(2n+1)(3n^2+3n-1))/30+3(n^2(n+1)^2)/4+5(n(n+1)(2n+1))/6+2(n(n+1))/2#

= #1/30(n^2+n)(2n+1)(3n^2+3n-1)+3/4n^2(n+1)^2+5/6n(n+1)(2n+1)+n(n+1)#

= #1/30(2n^3+3n^2+n)(3n^2+3n-1)+3/4n^2(n^2+2n+1)+5/6(n^2+1)(2n+1)+n(n+1)#

= #1/30(6n^5+15n^4+10n^3-n)+3/4(n^4+2n^3+n^2)+5/6(2n^3+n^2+2n+1)+n^2+n#

= #1/5n^5+n^4(1/2+3/4)+n^3(1/3+3/2+5/3)+n^2(3/4+5/6+1)+n(-1/30+5/3+1)+5/6#

= #1/5n^5+5/4n^4+7/2n^3+31/12n^2+79/30n+5/6#