SOLUTION: How can we prove the equation below? 1*1!+2*2!+3*3!+...+n*n!=(n+1)!-1

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Question 957846: How can we prove the equation below?
1*1!+2*2!+3*3!+...+n*n!=(n+1)!-1
Answer by amarjeeth123(569) About Me  (Show Source):
You can put this solution on YOUR website!
We can prove this equation using mathematical induction.
Let n=1
LHS=1*1!=1
RHS=2!-1=2-1=1
LHS=RHS
The equation holds true for n=1
We will assume that the equation holds true for n=k
1*1!+2*2!+.........+k*k!=(k+1)!-1......equation 1
For n=k+1 we have
LHS=1*1!+2*2!+.........+k*k!+(k+1)*(k+1)!
=(k+1)!-1+(k+1)*(k+1)!
=(k+1)!(k+1+1)-1
=(k+2)(k+1)!-1
=RHS
Therefore the equation holds true for all N.