Question

Can someone solve this by mathematical induction? ​

Answer 1

Answer:

Step-by-step explanation:

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