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I have problems proving this equality: $$1+\frac{n}{1!}+\frac{n*(n-1)}{2!}+\frac{n*(n-1)*(n-2)}{3!}+...+\frac{n*(n-1)...3*2}{(n-1)!}+\frac{n!}{n!}=2^n$$ Tried various options in inductive step of separating each addends that didn't help much.

EDIT: Any other way of solving this other then using Newton's binomial theorem ? I'm looking for a basic induction.

domdrag
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3 Answers3

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Hints:

$$\begin{align*}&\text{ Your sum is simply}\;\;\sum_{k=0}^n\binom nk=\sum_{k=0}^n\binom nk 1^k\cdot 1^{n-k}\\{}\\ &\text{ Newton's Binomial Theorem:}\;\;(a+b)^n=\sum_{k=0}^n\binom nk a^kb^{n-k}\end{align*}$$

DonAntonio
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  • I see.... May I ask for an induction that doesnt require that ? Im looking for a basic induction . – domdrag Apr 02 '17 at 19:10
  • @JamesGroon "Basic induction" is going to be, I'm afraid, a very cumbersome, lengthy thing. You can do induction to prove NBT, but also that is lengthy and uses combinatorial equalities. The easiest proof, perhaps, is the combinatorial one (also with some induction, though), but for that you'd need to know the meaning of $;2^n;$ ... – DonAntonio Apr 02 '17 at 19:14
  • Let's just hope there is someone here who is willingly to do this .Cuz there must be a way , the way I didn't see and I'm extremely curious in seeing that induction. I wanna see the "trick" . – domdrag Apr 02 '17 at 19:18
  • @JamesGroon Good luck with that. – DonAntonio Apr 02 '17 at 19:18
  • Do I need to be afraid of closing this topic ? – domdrag Apr 02 '17 at 19:19
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    @JamesGroon No, why close it? First, you already got an answer. Second, it is not a bad question though it'd require a lot of heavy work to do it as you want. Wait, perhaps there's some willing to do it... – DonAntonio Apr 02 '17 at 19:20
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First, notice that for all k $n \choose{k}$ = $\frac{n!}{(n-k)!*k!}$ = $\frac{n*(n-1)*..*(n-k+1)}{k!}$ So the left side of the equation can be written as $\sum_{k=1}^{n}{n \choose{k}}$

Now try to use ${n-1\choose{k-1}}+{n-1\choose k}={n\choose k}$ to split the sum.

The equality is proven here:Binomial coefficient proof for ${n\choose m-1}+{n\choose m}={n+1\choose m}$ to split the sum.

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Blueyedaisy
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  • Please see what I have written in "Edit" . I'm looking for a basic induction , simply separating addends ... – domdrag Apr 02 '17 at 19:23
  • This is exactly what it is. You have $2^n = 22^{n-1}$ and from induction $22^{n-1} = 2\sum_{k=1}^{n}{n \choose{k}}$ so you want to split the sum $ 2\sum_{k=1}^{n}{n \choose{k}}$ for which you can use the equality I wrote you. – Blueyedaisy Apr 02 '17 at 19:28
  • Basic meaning using standard way . There is absolutely no difference between using this or Newton's theorem . If you like , let's say I don't want any shortcut . I need hardcore induction. – domdrag Apr 02 '17 at 19:33
  • Ok I got you now. I would still try to separate the sums, just in a "dirtier" way: $\frac{n(n-1)..(n-k+1)}{k!} + \frac{n(n-1)..(n-k+2)}{k-1!} = \frac{n(n-1)..(n-k+2)}{k!}(n-k+1+ k)=$ $ \frac{n(n-1)..(n-k+2)}{k!}(n+1) = \frac{(n+1)n..*((n+1)-k+1)}{k!}$ , hope this is more of a help. Good luck. – Blueyedaisy Apr 02 '17 at 20:08
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This looks like you're trying to show that: $$ S_{n} =\sum_{j = 0}^{n} \dfrac{n!}{(n - j)!j!} = 2^{n} $$ each iteration of this sum is exactly the formula for choosing $j$ objects from a collection of $n$ objects where order doesn't matter (combinations) This is also the formula for the number subsets of a power set with $n$ elements and the way to find the sum of the $n$th row of Pascal's triangle.

Here is a Proof of the cardinality of a Power-set Using Induction (proof 3) that sketches the steps you need to take

Edit:

If you want to get there strictly by manipulating what you've been given I'd start by realizing that the first term (obviously) and last term of your sum are equal to 1, that the second and penultimate term are equal to $n$, that this is a convergent sum (and so rearrangeable) and that if we rewrite these pieces of my sum as $2, \textrm{ and } 2n$ ,respectively and that I can factor out $n(n-1)(n-2)$ from each of my remaining terms and that this number must be even (in other words I'm going to be able to fact out 2 from all of my terms)

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j.mac
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  • Need basic induction . Hardcore if you like . No shortcuts , no using theorems. Just using tricks to split the addends in inductive step. – domdrag Apr 02 '17 at 19:37