Claim: $-1+2+5+8+...+(3n-4) = \frac{n}{2}(3-5n)$
Base:
$3(1)-4=-1$
$\frac{1}{2}(3-5(1))=-1\,\,$
Induction:
$-1+2+5+8+...+(3k-4)+(3(k+1)-4) = \frac{k+1}{2}(3-5(k+1))$
$\frac{k}{2}(3-5(k))+[(3(k+1)-4] = \frac{k+1}{2}(3-5(k+1)) $#by induction hypothesis
$\frac{3k-k5^2}{2}+\frac{6k-2}{2} = \frac{(k+1)(-5k-2)}{2}$
$-5k^2+9k-2 = -5k^2-7k-2$#false
On
Assuming the book does not contain any typos in this excercise, your proof goes the right way, but it would be even better if you have added some comments in the induction step (and don't write down equations for which you don't know yet whether they hold or not):
We assume that the claim holds for $k$, i. e. $$-1 + 2 + 3 + \cdots + (3k - 4) = \frac{k}{2}(3 - 5k)$$ and try to prove that $$-1+2+5+8+\cdots+(3k-4)+(3(k+1)-4) = \frac{k+1}{2}(3-5(k+1))\text{.}$$
By induction hypothesis, LHS can be simplified to $$\frac{k}{2}(3 - 5k) + (3(k + 1) - 4)\text{,}$$ so we have to check if $$\frac{k}{2}(3 - 5k) + (3(k + 1) - 4) = \frac{k+1}{2}(3 - 5(k+1))\text{.}$$
Then, simplify LHS and RHS separatly and find out that the claim is incorrect.
Your prove looks fine to me. I believe it would be easier to transform one side at a time, but it's just writing way.
You proved that implication has incorrect consequent for true antecedent, so your implication isn't fulfilled and you can't conclude on the basis of mathematical induction.
If you had to prove for $(\forall n \in \mathbb{N}^+)(\sum_{i=1}^n(3n-4) = \frac{n}{2}(3-5n))$ you can also show that it's incorrect for $n=2$ for example.