Find the remainder when $\sum_{n=1}^{2015}{n^2\times2^n}$is divided by 23.

Since $2^{11}\equiv 1 \pmod{23}$ then the following holds for integers $q,r\geq 0$:

$$(q23+r)^2 \cdot 2^{q23+r} \equiv r^2 \cdot 2^{q+r} \pmod{23}$$ Therefore: $$ \sum_{r=0}^{22} (q23+r)^2 \cdot 2^{q23+r} \equiv 2^q \sum_{r=0}^{22} r^2 \cdot 2^{r} \pmod{23}$$ $$ \sum_{q=0}^{87} \sum_{r=0}^{22} (q23+r)^2 \cdot 2^{q23+r} \equiv (\sum_{q=0}^{87} 2^q) \sum_{r=0}^{22} r^2 \cdot 2^{r} \pmod{23}$$ Since $\sum_{q=0}^{87} 2^q \equiv 0 \pmod{23}$ (why?) we get $\sum_{n=0}^{88 \cdot 23 -1} n^2 2^n \equiv 0 \pmod{23}$. Does it help? ($88 \cdot 23 -1= 2023$ is not so far from $2015$ ... )