3

It is asked to be prove:

$$\forall{n}\in{N}:(n+1)(n+2)(n+3)...(n+n)=2^n\cdot1\cdot3\cdot5...\cdot(2n-1)$$

1 Step

p(n) is assumed to be true for n=1

$$(n+1)(n+2)(n+3)...(n+n)=2^n\cdot1\cdot3\cdot5...\cdot(2n-1)$$

Meaning that it is only consider the first term on the left, and only the $2^n\cdot1$ consider in the right hand.

$$(n+1)=2^n\cdot1$$

$$2=2 \square$$

But,

Question

What happens with n=3 for example?

My guess:

a:

$$(n+1)(n+2)(n+3)=2^n\cdot1\cdot3\cdot5$$ $$(3+1)(3+2)(3+3)=2^3\cdot1\cdot3\cdot5$$ $$120=120?$$

Step 2

Assuming that the proposition is also valid for p(k)

$$(k+1)(k+2)(k+3)...(k+k)=2^k\cdot1\cdot3\cdot5...\cdot(2k-1)$$

Step 3

Making the induction of validity for k+1 to also be true.

$$(k+1)(k+2)(k+3)...(k+k)+\boxed{(k+1)+(k+1)}=2^k\cdot1\cdot3\cdot5...\cdot(2k-1)+\boxed{(k+1)+(k+1)}$$

$$(k+1)(k+2)(k+3)...(k+k)+\boxed{(2k+2)}=2^k\cdot1\cdot3\cdot5...\cdot(2k-1)+\boxed{(2k+2)}$$

Question

Is this the right substation?

Haizum Skallah
  • 653

2 Answers2

1

(a) is the correct way of thinking about it. You said it yourself, you are looking at the $n=3$ case, so you have to plug that into the given equation. (b) is incorrect because $n$ to be a range of numbers ($n=1,2,3$) when you need it to just be one ($n=3$).

-Edit

You seem to have gotten rid of the (b) option. So then, yes that is the correct way of thinking about it for $n = 3$.

kbrose
  • 286
1

Hint: Here is what you want to prove. That is, when you replace $k$ with $k+1$, you obtain the following equation for $p(k+1)$:

$$ ((k+1)+1)((k+1)+2)((k+1)+3)\ldots((k+1)+(k+1))\\ =\\ 2^{k+1}\cdot1\cdot3\cdot5\cdot\ldots\cdot(2k-1)\cdot(2(k+1)-1) $$

Simplifying it a bit, this is equivalent to:

$$ (k+2)(k+3)(k+4)\ldots(2k+2) = 2^{k+1}\cdot1\cdot3\cdot5\cdot\ldots\cdot(2k-1)\cdot(2k+1) $$

Hopefully you can see how to finish off the induction now.

Adriano
  • 41,576