<s>[INST] 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.
 [/INST] For the inductive step the computations would be as follows:
$$
\begin{array}{rcl}
\sum_{k = 1}^{m+1} k^{2} 2^{k} & = & (m+1)^{2}2^{m+1} + \sum_{k = 1}^{m} k^{2}2^{k}\\
& = & (m+1)^{2}2^{m+1} + (m^2 - 2m + 3)2^{m+1} - 6\\
& = & 2^{m+1} \left((m+1)^2+m^2 - 2m +3\right) - 6\\
& = & 2^{m+1} (2m^2 + 4) - 6\\
& = & 2^{m+2} (m^2 + 2) - 6 \\
& = & 2^{m+2} (m^2 + 2m + 1 -2m - 2 + 3) - 6\\
& = & \left( (m+1)^{2} - 2(m+1) + 3 \right) - 6.
\end{array}
$$
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