- #1
james.farrow
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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!
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!