Find the value of:
1 × 1! + 2 × 2! + 3 × 3! + ...+ n × n!
Answers
Answer:
1 × 1! + 2 × 2! + 3 × 3! + ...+ n × n! = (n + 1)! - 1
Step-by-step explanation:
1 × 1! + 2 × 2! + 3 × 3! + ...+ n × n!
∑ n * n!
= ∑ (n+1 -1 ) * n!
= ∑ (n + 1) n! - n!
= ∑ (n + 1) ! - n!
=( 2! - 1! )+ (3! - 2!) +.............................+(n! - (n-1)!) + (n+1)! - n!)
= - 1 + (n + 1)!
= (n + 1)! - 1
1 × 1! + 2 × 2! + 3 × 3! + ...+ n × n! = (n + 1)! - 1
Answer:
1 × 1! + 2 × 2! + 3 × 3! + ...+ n × n! = (n + 1)! - 1
Step-by-step explanation:
In this question
we need to find the
1 × 1! + 2 × 2! + 3 × 3! + ...+ n × n!
Therefore, According to the question
∑ n × n!
= ∑ (n+1 -1 ) × n!
= ∑ (n + 1) n! - n!
= ∑ (n + 1) ! - n!
=( 2! - 1! )+ (3! - 2!) +.............................+(n! - (n-1)!) + (n+1)! - n!)
2! and -2! cancels similarly all the terms cancels out except
= - 1 + (n + 1)!
= (n + 1)! - 1
1 × 1! + 2 × 2! + 3 × 3! + ...+ n × n! = (n + 1)! - 1