I can't possibly reproduce all the algebra here, but it's doable - proof by induction.
Show it's true for N = 1 (clearly is)
Assume true for any N:
sum(k=1 to N) k(3^k-1) = 1/4[(3^N).(2N-1)+1]
then show that it must be true for N+1:
sum(k=1 to N+1) k(3^k-1) = 1/4[(3^(N+1)).(2(N+1)-1)+1]
Proof:
sum(k=1 to N+1) k(3^k-1) = 1/4[(3^N).(2N-1)+1] + (N+1)(3^N)
that last term is what's added to the original sum when k=N+1
So keep working on that expression (on the right hand side) and it DOES simplify to:
1/4[(3^(N+1)).(2(N+1)-1)+1]
QED.