algebra precalculus - Point out my fallacy, in sequence and series.

The sum of the first $n$--terms of the series $1^2+2\cdot2^2+3^2+2\cdot4^2+\cdots$ is $\dfrac{n(n+1)^2}{2}$, when $n$ is even. When $n$ is odd, the sum is?

I got the correct answer when is replaced $n\rightarrow (n+1)$ to make above valid for odd, but when I tried the different approach then something following had happened.

For $n$ even, last term $=n$ which is even and term before it $=n-1$ which is odd. Clubbing all odds and evens separately as follows:

$\big(1^2+3^2+\cdots +(n-1)^2\big)+2\big(2^2+4^2+\cdots+n^2\big)=\dfrac{n(n+1)^2}{2}\tag{1}$

For $n$ odd, last term $=n$ which is odd and term before it $=n-1$ which is even. Clubbing all odds and evens separately as follows:

$\big(1^2+3^2+\cdots +n^2\big)+2\big(2^2+4^2+\cdots+(n-1)^2\big)\tag*{}$

$=\big(1^2+3^2+\cdots +(n-1)^2\big)+2\big(2^2+4^2+\cdots+n^2\big)-n^2+(n-1)\tag*{}$

From equation $(1)$

$=\dfrac{n(n+1)^2}{2}-n^2+(n-1)\tag*{}$

And answer given is: $\dfrac{n^2(n+1)}{2}$

please help.