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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

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yep, check and check again, i also try reverse it but less the -(k+1),headache o, not understand –  Kai Nov 14 '12 at 17:13
    
What is the book? –  Julian Kuelshammer Nov 14 '12 at 17:35
    
Discrete and Combinatorial Mathematics by Ralph P. Grimaldi. Btw, i figure out it from the advise given below, thank all of u –  Kai Nov 14 '12 at 17:39
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3 Answers

up vote 2 down vote accepted

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 $$(k+1)!+(k+1)\cdot(k+1)!=(1+k+1)(k+1)!=(k+2)(k+1)!=(k+2)!$$

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thx, i take my time to digest this –  Kai Nov 14 '12 at 17:20
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Hint

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

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thx, try my best to understand –  Kai Nov 14 '12 at 17:21
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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...

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thx, i getting the big picture –  Kai Nov 14 '12 at 17:29
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