
Bobosharif S. answered 08/24/18
Tutor
4.4
(32)
PhD in Math, MS's in Calulus
1/[6k(k+1)(2k+1)+(k+1)2]=
=(k+1)[k(2k+1)/6+k+1)]=
((k+1)/6)[2k2+k+6k+6]= ((k+1)/6)[2k2+7k+6]=
((k+1)/6)(2k+3)(k+2)=
1/6(k+1)(k+2)(2k+3)=
Note: 2k2+7k+6=2(k+3/2)(k+2)=(2k+3)(k+2)
Vivien C.
08/24/18