Sum of the square of natural numbers - Mathematical Proof

Question

I'm trying to decipher why I have been marked down for this answer, is there a way in which this could be improved?

Question: Prove by induction the claim that the sum of the squares of the first $n$ natural numbers is equal to $(2n^3 + 3n^2 + n)/6$.

My workings:

$2n^3 + 3n^2 + n)/6 $

Base case: $n = 1$

$( 2(1)^3 + 3(1)^2 + 1 ) /6 $

$( 2 + 3 + 1 ) / 6 $

$6 / 6 = 1$ (true)

Inductive Hypothesis:
$n = k$, where $n \in \mathbb{N}$

Inductive Proof:
$n = k + 1?$

$[ 2 (k + 1)^3 + 3 (k + 1)^2 + (k + 1) ] / 6 $

$[ 2 (k^3 + 3k^2 + 3k + 1) + [ 3 (k^2 + 2k + 1) ] + (k+1) ] / 6 $

$(2k^3 + 6k^2 + 6k + 2) + (3k^2 + 6k + 3) ] + (k+1) ] / 6 $

$(2k^3 + 9k^2 + 13k + 6) / 6 $

Then using
 $(2k^3 + 3k^2 + k) / 6) + (k + 1)^2$

$= (2k^3 + 9k^2 + 13k + 6) $

Answer 1

The idea is correct, but it is very poorly written. The final part should be\begin{align}\frac{2 (k + 1)^3 + 3 (k + 1)^2 + (k + 1)}6&=\frac{2 (k^3 + 3k^2 + 3k + 1) + 3 (k^2 + 2k + 1) + (k+1)}6\\&=\frac{2k^3 + 6k^2 + 6k + 2 + 3k^2 + 6k + 3 + k+1}6\\&=\frac{2k^ 3 + 9k^ 2 + 13k + 6}6\\&=\frac{2k^ 3 + 3k^ 2 + k}6+(k+1)^2\\&=1^2+2^2+\cdots+k^2+(k+1)^2,\end{align}by the induction hypothesis.

Answer 2

You have to show the other way around:

Inductive hypothesis: $$n=k \Rightarrow 1^2+2^2+\cdot +k^2=\frac{2k^3+3k^2+k}{6}.$$

Inductive step: $$n=k+1 \Rightarrow 1^2+2^2+\cdot +k^2+(k+1)^2=\frac{2k^3+3k^2+k}{6}+(k+1)^2=\\ \frac{2k^3+3k^2+k+6k^2+12k+6}{6}=\frac{2(k^3+3k^2+3k+1)+3(k^2+2k+1)+(k+1)}{6}=\\ \frac{2(k+1)^3+3(k+1)^2+(k+1)}{6}.$$