Use induction to prove that for all integers n >=1, 1 x 1 ! + 2 x 2 ! + - - - n x n ! = ( n + l ) ! - 1?

Answer 1

We wish to show that:
[k=1, n]∑kk! = (n+1)! - 1

Clearly, this is true for n=1, since 1*1! = 2! - 1 = 1. Suppose it is true for some n. Then for n+1:
[k=1, n+1]∑kk!
(n+1)*(n+1)! + [k=1, n]∑kk!
(n+1)*(n+1)! + (n+1)! - 1
(n+2)*(n+1)! - 1
(n+2)! - 1

Which is what we would get if we substituted n+1 for n in our formula. So by induction, the formula holds for all n. Q.E.D.

Answer 2

base case n =1 then 1*1! =1 = 2! -1
induction hypothesis: assume
for k >= 1 1x1! + ... + kxk! = (k+1)! -1
add to both sides of the equation (k+1)x(k+1)!
getting
1x1! + ... + kxk! + (k+1)x(k+1)! = (k+1)! -1 (k+1)x(k+1)! we have the left side onf the equation done.
now (k+1)! -1 (k+1)x(k+1)! = (k + 1)!(1 + k+1 ) -1 = (k+1)!(k+1 + 1) -1= ((k+1) + 1)! - 1 Q.E.D.

Answer 3

not an answer, but im intrigued, what is this about ,, not very good without a calculator but still like a puzzle