# Thread: [SOLVED] prove by mathematical induction...it's not true

1. ## [SOLVED] prove by mathematical induction...it's not true

Prove by mathematical induction that 2+5+8+...+(3k-1) = 1/2n(3n+1)

k^2 = 2+5+8+...+(3k-1)
(k+1)^2 = 2+5+8+...+ 3(k+1) + [3(k+1)-1]
(k+1)^2 = 2+5+8+...+ 3(k+1) + 3k + 2
(k+1)^2 = k^2+3k+2
k^2 + 2k +1 =/ k^2 +3k +2

Did I do something wrong or is the problem not true?

2. I assume you mean $2+5+8+....+(3n-1)=\frac{1}{2}n(3n+1)$

$2+5+8+.....+(3k-1)+(3k+2)=\frac{1}{2}k(3k+1)+(3k+2)$

$2+5+8+....+(3k-1)+(3k+2)=\frac{1}{2}(k+1)(3(k+1)+1)$

It is shown.

You can put in the usual details.

3. Originally Posted by galactus
I assume you mean $2+5+8+....+(3n-1)=\frac{1}{2}n(3n+1)$

$2+5+8+.....+(3k-1)+(3k+2)=\frac{1}{2}k(3k+1)+(3k+2)$

$2+5+8+....+(3k-1)+(3k+2)=\frac{1}{2}(k+1)(3(k+1)+1)$

It is shown.

You can put in the usual details.
is the right side of the equation:
1.5k^2 + 3.5k +2??

4. Yes, it is. But, to show induction nicely, you may want to keep it in the k+1 form I showed you.

5. omg, im so lost! does 2+5+8+...(3k-1) = k^2?

so should it be:
K^2+3k+2 = (3k+4)(k+1)/2

This book makes no freaking sense to me!!