# Tricky maths 2

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]
19:16 Mon 30th Apr 2012
 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) =... 19:35 Mon 30th Apr 2012 Go To Best Answer

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 Tridy you don't have to start a new post, just correct the old one! 19:16 Mon 30th Apr 2012 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. 19:35 Mon 30th Apr 2012

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## Tricky maths

Can anyone help with this? Show that the sum between 1 and k of k.3^(3-1) =1/4[(3^k).(2k-1)+1]....