Tricky maths 2

[Tridymite]

Sorry, there wsa an error.

Show that the sum for 1 to k of k.3^(k-1)=1/4[(3^k).(2k-1)+1]

[boxtops]

Tridy you don't have to start a new post, just correct the old one!

[dr b]

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.