Prove Math Induction: 1*3^1 + 2*3^2 + 3*3^3 +...+ n*3^n = 1/4(2n-1)3^n+1 + 3/4

[james.farrow]

Use mathematical induction to prove

1*3^1 + 2*3^2 + 3*3^3 + ... + n*3^n = 1/4(2n -1)3^n+1 +3/4

Let p(n) be variable proposition 1/4(2n -1)3^n+1 + 3/4

Is p(1) true?

1/4(2*1 - 1)3^1+1 + 3/4 = 3 & 1*3^1 = 3 so p(1) is true.

Now assume p(k) is true, that is 1/4(2k - 1)3^k+1 + 3/4 is true.

Now prove p(k + 1) is true.

If p(k + 1) is true we should have 1/4(2(k+1) - 1)3^(k+1) +1 + 3/4
which is

1/4(2k + 1)3^k+2 +3/4

Now to proof.

We have to prove that p(k) + p(k+1) = 1/4(2k + 1)3^k+2 + 3/4

1/4(2k + 1)3^k+1 + 3/4 + (K+1)3^(k+1)

Multiplying by 4

(2k + 1)3^k+1 + 3 + 4(k+1)3^(k+1)

This is where I get bogged down? Can anyone help please!

 

[danago]

...
1/4(2k + 1)3^k+1 + 3/4 + (K+1)3^(k+1)
...

Should that be 2k - 1?

Anyway, after fixing that, try taking out a common factor of 3^(k+1). See if you can go from there.

 

[james.farrow]

Yeah you're right! lol

After some head scratching and crossings out I think I've cracked it!

Anyway after fixing my error it simplifies to

18k*3^k + 9*3^k +3

(2k + 1)9*3^k + 3

But 9 = 3^2 so we have

(2k + 1)3^2*3^k + 3
(2k + 1)3^k+2 +3

Now multiply by 1/4 and there is the answer!

1/4(2k + 1)3^k+2 +3/4


I think!

Many thanks for your help.

James