<s>[INST] Prove that for every $k\geq 1$, $1\cdot 3\cdot(1!)^2+....+k\cdot(k+2)\cdot(k!)^2=((k+1)!)^2-1$ Prove that for every $k\ge1,$ 
$\displaystyle1\cdot3\cdot(1!)^2+2\cdot4\cdot(2!)^2+....+k\cdot(k+2)\cdot(k!)^2=((k+1)!)^2-1$
I start off with considering induction, but I can't quite get the left side to match the right side. Maybe there is something I'm not seeing or a calculation I have wrong. Any kind of help would be great!
$\displaystyle((n+1)!)^2-1+(n+1)(n+3)((n+1)!)^2=((n+2)!)^2-1$
LHS:
$\displaystyle((n+1)!)^2[(n+1)(n+3)-1]$
$\displaystyle((n+1)!)^2[(n+2)^2-2]$
 [/INST] Let 
$\displaystyle f(k): 1\cdot3\cdot(1!)^2+2\cdot4\cdot(2!)^2+....+k\cdot(k+2)\cdot(k!)^2=((k+1)!)^2-1$  holds true for $k=m$
$\displaystyle\implies 1\cdot3\cdot(1!)^2+2\cdot4\cdot(2!)^2+....+m\cdot(m+2)\cdot(m!)^2=\{(m+1)!\}^2-1 $
For $k=m+1,$
$\displaystyle\implies 1\cdot3\cdot(1!)^2+2\cdot4\cdot(2!)^2+....+m\cdot(m+2)\cdot(m!)^2+\underbrace{(m+1)\{(m+1)+2\}\{(m+1)!\}^2}$
$\displaystyle=\{(m+1)!\}^2-1+\underbrace{(m+1)\{(m+1)+2\}\{(m+1)!\}^2} $
$\displaystyle=\{(m+1)!\}^2\{1+(m+1)(m+3)\}-1$
$\displaystyle=\{(m+1)!\}^2(m+2)^2-1$ as $1+(m+1)(m+3)=(m+2)^2$
$\displaystyle=\{(m+2)\cdot (m+1)!\}^2-1$
$\displaystyle=\{(m+2)!\}^2-1$
So, $f(k)$ will hold true for $k=m+1,$ if it holds true for $k=m$
Now establish the base  case i.e., $k=1$
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