Prove, by mathematical induction, or otherwise, that 1*1!+2*2!+3*3!+…+n*n!=(n+1)!-1
To prove the identity 1*1! + 2*2! + 3*3! + ... + n*n! = (n+1)! - 1 using mathematical induction, we need to follow these steps:
Step 1: Base case
First, we need to prove that the identity holds for the smallest possible value of n. Let's take n = 1 and substitute it into the equation:
1*1! = (1+1)! - 1
1 = 2! - 1
1 = 2 - 1
1 = 1
As the equation holds true for n = 1, we have established the base case.
Step 2: Inductive hypothesis
Assume that the equation holds true for some value k. That is, assume 1*1! + 2*2! + 3*3! + ... + k*k! = (k+1)! - 1.
Step 3: Inductive step
We now need to prove that the equation also holds for k+1. Let's consider:
1*1! + 2*2! + 3*3! + ... + k*k! + (k+1)*(k+1)! = (k+1+1)! - 1
We can rewrite the left-hand side of the equation as:
(k+1)! - 1 + (k+1)*(k+1)! = (k+1)! + (k+1)*(k+1)! - 1
Now, we can factor out (k+1)! from the two terms:
[(k+1)! + (k+1)*(k+1)!] = [(k+1)! * (1 + (k+1))] - 1
Simplifying the right-hand side:
[(k+1)! + (k+1)*(k+1)!] = [(k+1)! * (k+2)] - 1
We can rewrite this as:
[(k+1)! * (k+2)] - 1 = (k+2)! - 1
This shows that the equation holds for k+1, completing the inductive step.
Therefore, by mathematical induction, we have proved that 1*1! + 2*2! + 3*3! + ... + n*n! = (n+1)! - 1.