The base case is n=1, so S1=12(2.12-1)=1 which is T1. Tn=(2n-1)3.
By the proposed formula Sn+1=Sn+Tn+1=n2(2n2-1)+(2n+1)3. Sn=sum of the first n terms, Tn is the nth term.
S2=4(8-1)=28, S3=9(18-1)=153. From the series, S2=13+33=28, S3=13+33+53=153.
Tn+1=(2(n+1)-1)3=(2n+1)3=8n3+12n2+6n+1.
Sn+1=2n4-n2+8n3+12n2+6n+1=2n4+8n3+11n2+6n+1.
2(n+1)4-(n+1)2=2(n4+4n3+6n2+4n+1)-(n2+2n+1)=2n4+8n3+11n2+6n+1.
So Sn+1=Sn+Tn+1, and the formula Sn=n2(2n2-1) applies QED.