Let n be a positive natural number , n≥0, then. n∑i=0i⋅i!=(n+1)!−1
Here is my attempt.I'm not going to write the base case because I understand that part.
Assuming n∑i=0i⋅i!=(n+1)!−1 is true. We wish to show.
n+1∑i=0i⋅i!=(n+2)!−1. Thus.
n+1∑i=0i⋅i!=(n∑i=0i⋅i!) This is where I get stuck.
Answer
Your last step should read k+1∑i=0(i⋅i!)=k∑i=0(i⋅i!)+(k+1)(k+1)!
Now use your induction assumption to get k+1∑i=0(i⋅i!)=(k+1)!−1+(k+1)(k+1)!=(k+1)!(k+2)−1=(k+2)!−1
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