Factorial number

Question

Can anybody explain how $$ (k+2)(k+1)!-1 = (k+2)!-1 $$ also how $$ (k+1)!-1+(k+1)(k+1)! = [1+(k+1)](k+1)!-1 $$ my book show this example but i can't understand how. I also try google it but cannot write it in word, sad

Answer 1

The general recursive definition of factorial is $$0!=1$$ $$(n+1)!=(n+1)\cdot n!\quad\text{for }n\ge0$$ Thus especially $$(k+2)!=(k+2)\cdot (k+1)!$$ Also, by distributivity $$(k+1)!+(k+1)\cdot(k+1)!=(1+k+1)(k+1)!=(k+2)(k+1)!=(k+2)!$$

Answer 2

Hint

Use $k! = k(k-1)!$. The second one is just factoring the right terms out.

Answer 3

First, I assume you mean $(k+1)! \times (k+2) - 1 = (k+2)! - 1.$

To see that they are, just apply the definition of factorial. For example: $3! = 1 \times 2 \times 3.$ It follows that $[3!] \times 4 = [1 \times 2 \times 3] \times 4 = 4!$. In general:

$$(k+1)! \times (k+2) = [1 \times 2 \times \cdots \times k \times (k+1)] \times (k+2) = (k+2)!$$

If $(k+1)! \times (k+2) = (k+2)!$ then $(k+1)! \times (k+2) -1 = (k+2)!-1.$

For your second question, we can add $1$ to both sides to give

$$(k+1)!+(k+1)\times (k+1)! = [1+(k+1)](k+1)!$$

There is a common factor of $(k+1)!$ on the left and so:

$$(k+1)![1+(k+1)] = [1+(k+1)](k+1)!$$

Done...