Induction Proof $k^2 \times 2^k$ I need help on this proof. I am not able to do it after setting m=m+1.
Prove by induction on n that sum of $k^2 \times 2^k$ from $k=1$ to $n$ is equal to $(n^2-2n+3) \times 2^{n+1}-6$
Base case:
Let $k=1$ so L.H.S side is $2$
Let $n=1$ so R.H.S side is $2$
Inductive hypothesis:
Let $n=m$ so $(m^2-2m+3) \times 2^{m+1}-6$
Proof:
Let $n=m+1$ so prove that $((m+1)^2-2(m+1)+3) \times 2^{m+2}-6=(m^2-2m+3) \times 2^{m+1}-6$
But I am not able to prove that they are equal.
 A: For the inductive step the computations would be as follows:
$$
\begin{array}{rcl}
\sum_{k = 1}^{m+1} k^{2} 2^{k} & = & (m+1)^{2}2^{m+1} + \sum_{k = 1}^{m} k^{2}2^{k}\\
& = & (m+1)^{2}2^{m+1} + (m^2 - 2m + 3)2^{m+1} - 6\\
& = & 2^{m+1} \left((m+1)^2+m^2 - 2m +3\right) - 6\\
& = & 2^{m+1} (2m^2 + 4) - 6\\
& = & 2^{m+2} (m^2 + 2) - 6 \\
& = & 2^{m+2} (m^2 + 2m + 1 -2m - 2 + 3) - 6\\
& = & \left( (m+1)^{2} - 2(m+1) + 3 \right) - 6.
\end{array}
$$
A: The induction hypothesis is that
$$\tag1\sum_{k=1}^m k^22^k=(m^2-2m+3)2^{m+1}-6$$
Now, 
$$\sum_{k=1}^{m+1}k^22^k=\sum_{k=1}^mk^22^k+(m+1)^22^{m+1}$$ 
Inductive Step:
From $(1)$,
$$\sum_{k=1}^{m+1}k^22^k=\sum_{k=1}^mk^22^k+(m+1)^22^{m+1}=(m^2-2m+3)2^{m+1}-6+(m+1)^22^{m+1}=2^{m+1}(m^2-2m+3+m^2+2m+1)-6=2^{m+1}(2m^2+4)-6=2^{m+2}(m^2+2)-6$$ 
Moreover,
$$(m+1)^2-2(m+1)+3=m^2+2m+1-2m-2+3=m^2+2$$
Hence,
$$\sum_{k=1}^{m+1}k^22^k=2^{m+2}(m^2+2)-6=2^{m+2}((m+1)^2-2(m+1)+3)-6$$
Thus, given the statement is true for $m$, it is true for $m+1$.
By the principle of mathematical induction, it is true for all $m\in\mathbb N$.
