Answer:
Below.
Explanation:
First check if its true for n = 1:-
∑1(1!) = 1*1 = 1
(n + 1)! - 1 = 2! - 1 = 2 - 1
= 1.
So its true for n = 1.
Suppose it's true for a specific value n = k, then we have
∑k(k!) = (k + 1)! - 1
The (k + 1)th term is (k + 1)(k + 1)!
So ∑k+1(k+1!) = (k + 1)! - 1 + (k + 1)(k + 1)!
= (k + 1)(k + 1)!- 1.
- this result is the same as the result for ∑k(k!) except that the k is replaced by k+1. So if its true for n = k then its also true for n = k+1.
We have shown that the formula is true for n = 1 so therefore it is true for 1, 2, 3, 4... (all whole numbers).
This concludes the proof.