Question

Can someone solve this by mathematical induction?
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Answer 1

Explanation:

`1) Initialisation:\\n=1\\1*1!=1\\(1+1)!-1=2-1=1\\True\\\\2) recursivity:\\\displaystyle \sum_(r=1)^n(r*r!)=(n+1)!-1\ is\ true\\\displaystyle \sum_(r=1)^(n+1)(r*r!)= \sum_(r=1)^(n)(r*r!)+(n+1)*(n+1)!\\\\=(n+1)!-1+(n+1)*(n+1)!\\=(n+1)!*(1+(n+1))-1\\=(n+1)!*(n+2)-1\\=(n+2)!-1\ thus\ true\ for\ n+1.\\`