Hey Everyone,
So here is the problem I am currently working on and I have a few questions with what I can factor our of the sigma notation in order to use the IH.
Base Case: $1$ is the base case $1 = 1$
IH: Basically Restate the Question
WTP: $n+1$ holds for this equation. How I understand that the right side of the equation ends up looking like $(n+2)(n+1)! - 1$. However I am confused about the left side of the equation.
Plugging n+1 into the sigma notation I get the highest term where $k = n+1$ to be $((n+1)\cdot (n+1)!)$. Now how do I go about factoring this out? I want it to look like the original LHS of the equation when $k = n$ so I can use my IH. So you can factor into $((n+1)\cdot (n+1)(n)!)$.
But where do I go from here? I am completely stuck and dont know how to factor further.
Thanks!
\begin{align*} \sum_{k=1}^{n+1}k(k!) &= \sum_{k=1}^n k(k!) + (n+1)(n+1)!\\& = (n+1)!-1+(n+1)(n+1)! \\&= (n+1)!(1+n+1)-1 \\&= (n+2)!-1.\end{align*}