Question

1x1! +2x2!+3x3!+4x4!+....+nxn!

Answer 1

What is this?

Answer 2

What do you want to do with this sum?

Answer 3

find the value

Answer 4

in terms of n

Answer 5

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

3 steps in induction:
a) Prove true for trivial case n = 1.
b) Assume true for some n = k; this is the Inductive Hypothesis IH.
c) Assuming IH, prove for n = k + 1.

a) Prove for n = 1: 1*1! = 1.

b) Assume

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

c) Now we want to show

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

LHS = [1*1! + 2*2! + 3*3! + .... + n*n!] + (n + 1)*(n + 1)!

= (n + 1)! - 1 + (n + 1)*(n + 1)! (replace [ ] by IH)

= (n + 1)![1 + n + 1] - 1

= (n + 1)!*(n + 2) - 1

= (n + 2)! - 1 = RHS,

since the recursive rule for factorials is (m + 1)*m! = (m + 1)!.

Answer 6

consider using the following:
\[n \cdot n! =(n+1-1) \cdot n!=(n+1) \cdot n!-1 \cdot n! \\ n \cdot n! =(n+1) n!-n!=(n+1)!-n! \\ n \cdot n!=(n+1)!-n!\]
you would rewrite each term
for example:
\[3 \cdot 3!=4!-3!\]