Tricky maths 2

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Show that the sum for 1 to k of k.3^(k-1)=1/4[(3^k).(2k-1)+1]
20:16 Mon 30th Apr 2012
 
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dr b
Best Answer
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) =...
20:35 Mon 30th Apr 2012 Go To Best Answer

1 to 2 of 2

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

1 to 2 of 2

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