# 1 into 1! + 2 into 2 ! + 3 into 3!+ ... + 10 into n factorial=(n+1)!-1. prove this

Dear Student,
```First we prove it's true for n=1

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

Now we assume it's true for n=k

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

We need to show that

(2)     1(1!)+2(2!)+3(3!)+...+(k+1)(k+1)! ≟ (k+2)!-1

We add (k+1)(k+1)! to both sides of (1)

(1)     1(1!)+2(2!)+3(3!)+...+k(k!)+(k+1)(k+1)! = (k+1)!-1+(k+1)(k+1)! =
= (k+1)!+(k+1)(k+1)!-1 =
= (k+1)![1+(k+1)]-1 =
= (k+1)![1+k+1]-1 =
= (k+1)!(k+2)-1 =
= (k+2)!-1

So the truth of (1) implies the truth of (2). So the induction is complete.
Regards
```

• 1
We use mathematical induction. In fhe basis step, for n = 1, the equation states that 1 · 1! = (1 + 1)! − 1, and this is true because both sides of the equation evaluate to 1. For the inductive step, we assume that 1 · 1! + 2 · 2! + · · · + k · k! = (k + 1)! − 1 for some positive integer k. We add (k + 1)(k + 1)! to the left hand side to find that 1 · 1! + 2 · 2! + · · · + (k + 1) · (k + 1)! = (k + 1)! − 1 + (k + 1)(k + 1)! The right hand side equals (k + 1)!(k + 2) − 1 = (k + 2)! − 1. This establishes the desired equation also for k + 1, and we are done by the principle of mathematical induction.We use mathematical induction. In fhe basis step, for n = 1, the equation states that 1 · 1! = (1 + 1)! − 1, and this is true because both sides of the equation evaluate to 1. For the inductive step, we assume that 1 · 1! + 2 · 2! + · · · + k · k! = (k + 1)! − 1 for some positive integer k. We add (k + 1)(k + 1)! to the left hand side to find that 1 · 1! + 2 · 2! + · · · + (k + 1) · (k + 1)! = (k + 1)! − 1 + (k + 1)(k + 1)! The right hand side equals (k + 1)!(k + 2) − 1 = (k + 2)! − 1. This establishes the desired equation also for k + 1, and we are done by the principle of mathematical induction.
• 2
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