Sn+1=Sn+(n+1)2n+1, where Sn is the sum to n terms.
The nth term is Tn=n(2n), T1=1(21)=2 so S1=S0+2, so S0=T0=0. The given series therefore is shown starting at T1, not T0. S2=S1+2.22=S1+8=10. By the proposed formula S1=2[1+1.(21)]=2×3=6. But S1=2, as we saw before. Therefore the formula is not correct. By the formula, S2=2[1+2.(22)]=18, whereas S2=T0+T1+T2=0+2+8=10. S3=2[1+3.(23)]=50, which should have been 34. If we compare the actual results with the proposed results we get differences of 6-2=4; 18-10=8; 50-34=16. So there's a pattern here because each difference is a power of 2.
We need to establish the correct formula. 2[1+n2n]-2n+1 seems to fit.
Sn=2+n2n+1-2n+1=2+2n+1(n-1)=2[1+(n-1)2n].
When n=0, S0=2[1-1]=0, S1=2[1+0]=2, S2=2[1+4]=10, S3=2[1+16]=34.
So, Sn+1=Sn+(n+1)2n+1=
2[1+(n-1)2n]+(n+1)2n+1=
2+n2n+1-2n+1+n2n+1+2n+1=2+2n2n+1=2[1+n2n+1]=Sn+1 by definition through the new formula. The base case was proved by using the formula to derive S0, so by induction Sn=2[1+(n-1)2n]. This formula works for integer n≥0, so it also works for natural numbers n, because the set of natural numbers is a subset of integers.