Question

Establish the following formula by principle of mathematical induction.
1(1!)+ 2(2!)+ 3(3!)+............................+n(n!) = (n+1)! - 1​

Answer 1

Answer:

Suppose the statement is true for some value of n (this assumption is the "inductive hypothesis").

We need to show that then the statement is also true for the next value, that is, for n+1.  In other words, we need to show that

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

Starting with the left hand side, we get

 1(1!) + 2(2!) + 3(3!) + ... + n(n!) + (n+1)(n+1)!

= (n+1)! - 1 + (n+1)(n+1)!                [ by the inductive hypothesis ]

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

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

= (n+2)! - 1

So, under the assumption that the statement holds for some value n, the statement then also holds for the next value n+1.

All that remains is to show that the statement is true for some starting value to kick off the whole induction.  Checking the statement for n=1, we see

   1(1!) = 1×1 = 1  = 2 - 1 = 2! - 1  = (1+1)! = 1.

So the statement holds for n=1, and so by induction, it holds for all natural numbers n.

Hope that helps.