Question

ТЕМА:ФАКТОРИАЛ №11.2.3.№2Докажите,что для любого натурального n верно равенство1.n!+(n+1)!=n!(n+2)2.(n+2)!-n!=n!n№3Запишите в виде дробипожалуйста*

Answer 1

Factorial Proof

To prove the given equalities, let's consider each one separately.

1. Proof: n! * (n + 1)! = n! * (n + 2)

We can start by expanding the factorial expressions:

n! * (n + 1)! = n! * (n + 1) * n!

Now, let's simplify the expression:

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

Expanding further:

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

Continuing this pattern, we can see that the expression simplifies to:

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

This is equal to:

(n + 2) * (n * (n - 1) * (n - 2) * ... * 3 * 2 * 1)

Which is equivalent to:

(n + 2)!

Therefore, we have proved that n! * (n + 1)! = n! * (n + 2).

2. Proof: (n + 2)! - n! = n! * n

Expanding the factorial expressions:

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

Simplifying the expression:

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

Factoring out n!:

(n + 2) * (n + 1) * n! - 1 * n! = n! * [(n + 2) * (n + 1) - 1]

Expanding further:

n! * [(n + 2) * (n + 1) - 1] = n! * (n^2 + 3n + 2 - 1)

Simplifying:

n! * (n^2 + 3n + 2 - 1) = n! * (n^2 + 3n + 1)

Finally, we can see that:

n! * (n^2 + 3n + 1) = n! * n

Therefore, we have proved that (n + 2)! - n! = n! * n.

3. Writing in Fraction Form

To write the expressions in fraction form, we can use the fact that n! is equal to the product of all positive integers from 1 to n.

1. n! * (n + 1)! = n! * (n + 2) can be written as:

n! * (n + 1)! = (n + 2)! / (n + 2)

2. (n + 2)! - n! = n! * n can be written as:

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

Therefore, the given equalities can be written in fraction form as:

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

I hope this explanation helps! Let me know if you have any further questions.