I can't seem to detect where I'm going wrong so maybe I've made a fundamental flaw in my derivation.
First I start with $$(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^k y^{n-k}$$
Then I differentiate with respect to x and multiply by x to get,
$$xn(x+y)^{n-1} = \sum_{k=0}^n k \binom{n}{k}x^k y^{n-k}$$
At this point I can simply plug in $x=p$ and $y=1-p$ to get the mean, $pn$ as usual.
I differentiate with respect to x again and then multiply by x, yielding,
$$x \frac{d}{dx} \left( xn(x+y)^{n-1} \right) = \sum_{k=0}^n k^2 \binom{n}{k} x^k y^{n-k}$$ $$x(n(x+y)^{n-1} + x n(n-1)(x+y)^{n-2}) = \sum_{k=0}^n k^2 \binom{n}{k} x^k y^{n-k}$$ $$xn(x+y)^{n-1} + x^2 n(n-1)(x+y)^{n-2} = \sum_{k=0}^n k^2 \binom{n}{k} x^k y^{n-k}$$
This formula should be true with x and y as independent variables. It's simply the relation between the right hand side to the mildly nicer left hand side.
Now plugging in $x=p$ and $y=1-p$ out comes what should be the variance,
$$pn(p+1-p)^{n-1} + p^2 n(n-1)(p+1-p)^{n-2} = \sum_{k=0}^n k^2 \binom{n}{k} p^k (1-p)^{n-k}$$ Which simplifies considerably to
$$pn+p^2 n^2-p^2n$$
However I know the variance of the binomial distribution is
$$pn -p^2n$$
So how did I accidentally acquire this extra $p^2n^2$ term, I can't seem to find the mistake in my derivation.
The sum $$\sum k^2\binom{n}{k}p^k (1-p)^{n-k}$$ is not $V(X)$ but $E(X^2)$. The variance is $$V(X) = E(X^2)- E(X)^2$$