Prove 1*1! + 2*2! + 3*3!+ ... + n*n! + (n+1)*(n+1)! = (n+2)! -1

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SUMMARY

The discussion centers on proving the equation 1*1! + 2*2! + 3*3! + ... + n*n! = (n+1)! - 1 using mathematical induction. The proof begins with a base case for n=1, confirming that 1=1 holds true. The inductive step assumes the equation is valid for n and aims to prove it for n+1, ultimately demonstrating that (n+2)! - 1 equals the sum of the previous terms plus (n+1)(n+1)!. This establishes the validity of the formula for all natural numbers n.

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Homework Statement



Prove that 1/2 P(2,1) + 2/3 P(3,2)+3/4 P(4,3)+ ... + n/(n+1) P(n+1,n) = (n+1)! - 1

Please help!

Homework Equations


P(n,r) = n!/(n-r)!

The Attempt at a Solution


The inequation can be simplified to:
1*1! + 2*2! + 3*3!+ ... + n*n! = (n+1)! - 1 (*)
Use the induction method:
1/Base case: n=1 -> 1=1 true
2/Inductive case: suppose (*) is true
Need to prove: 1*1! + 2*2! + 3*3!+ ... + (n+1)*(n+1)! = (n+2)! -1
 
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i think you;re pretty much there, so assume f(n) = (*) is true, then you need to show
f(n+1) = (n+2)! - 1 = f(n) + (n+1)(n+1)!

substituting for f(n) gives
(n+2)! - 1 = (n+1)! - 1 + (n+1)(n+1)!
 

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