Prove you are not a brainlet and do this exam question I just

A(1) holds as
[math] 1=\sum_{U\subseteq\{1\}}\prod_{j\in U}\frac{1}{j}=\frac{1}{1}=1\,.[/math]
someone else can do the rest

>>8878859
Explain the notation and I will be sure to solve it for you, as my IQ is 257.

>>8878876
ight well this anon just did the base case for n=1.
you can assume n=k, so prove n=k+1 is true using induction.

File: blonde-emma-stone-1280x800-celebrity-wallpaper.jpg (100KB, 1280x800px) Image search: [Google]

100KB, 1280x800px

You can rewrite it as

[math] \prod_{j=1}^{n} \left( 1+ \dfrac{1}{j} \right) = 1+n [/math]

For [math] n=1[/math], both sides are [math] 2[/math] and it holds.
The induction step is

[math] (1+n) \left( 1+ \dfrac{1}{n+1} \right) = (1+n) + 1 [/math]

Mostly, I reply to point out that the equation above generalizes to

[math] \prod_{k=a}^{b-1}\dfrac{B+k}{A+k} = \prod_{k=A}^{B-1}\dfrac{b+k}{a+k} [/math]

which I find extremely funky.

>>8878897
there is a lot of notation

>>8878859
[math]
\displaystyle \sum_{U \in \wp \{1,2,3\} } \prod_{k \in U} \frac{1}{k} \\
\displaystyle \sum_{k_1 \in \{0,1\}} \sum_{k_2 \in \{0,1\}} \cdots \sum_{k_n \in \{0,1\}} \frac{1}{1^{k_1}} \frac{1}{2^{k_2}} \cdots \frac{1}{n^{k_n}} \\
\displaystyle \sum_{k_1 \in \{0,1\}} \frac{1}{1^{k_1}} \sum_{k_2 \in \{0,1\}} \frac{1}{2^{k_2}} \cdots \sum_{k_n \in \{0,1\}} \frac{1}{n^{k_n}} \\
\displaystyle (1+\frac{1}{1}) (1+\frac{1}{2}) \cdots (1+\frac{1}{n}) \\
\displaystyle \frac{2 \cdot 3 \cdots (n+1)}{1 \cdot 2 \cdots n} \\
\displaystyle n+1
[/math]

File: 4164864684684.jpg (156KB, 2338x566px) Image search: [Google]

156KB, 2338x566px

Alright so... to be honest I don't know where some of you are headed with it... but here is a base case of n=3 to get a better understanding of the pattern here.

Once I saw a pattern it became easier for me on the exam.

>>8878952
yo thats fucking tricky

File: 546846468448864648.jpg (410KB, 2338x1700px) Image search: [Google]

410KB, 2338x1700px

>>8879015
And here is what I wrote for the k and then did some dumb shit for (k+1) and just hoped I made enough sense to get full credit.

>>8879015
I posted

[math] 1 + \sum_{U \subset \{1,...,n\} } \prod_{j \in U} a(n) = \prod_{j=1}^{n} \left( 1+ a(j) \right) [/math]

above already

Y U no accept?

>>8879063
If you are emma stone up there then I didn't understand your first line when you rewrote the equation.
Taking another look to see what you did.

>>8879107
The first line is the claim in OPs pic together with >>8879063

To see it, consider e.g.

(1+a)·(1+b)·(1+c) = 1 + a + b + c + a·b + a·c + a·c + a·b·c

The non-empty subsets of {1,2,3,..,n+1} break down as the subsets of {1,2,3,...,n}, plus those subsets again but with n+1 as an extra element, and the subset {n+1}.

So
[math]\begin{align} \displaystyle \sum_{U \in \wp \{1,2,3,...,n+1\} } \prod_{k \in U} \frac{1}{k} &= \sum_{U \in \wp \{1,2,3,...,n\} } \prod_{k \in U} \frac{1}{k} + \frac{1}{n+1} \sum_{U \in \wp \{1,2,3,...,n\} } \prod_{k \in U} \frac{1}{k} + \frac{1}{n+1} \\ &= n +\frac{1}{n+1}n + \frac{1}{n+1} \\ &=n + \frac{n+1}{n+1} \\ &=n+1\end{align}[/math]

OP I think this is really easy (if I understood it correctly).
Let's regard the right hand term as a function f(n).

to compete f one has to first find the set of sets which are in the equation.
What I mean is that for f(3) you first find the set: {1, 2, 3, 12, 13, 123}
I'll call this M(n)
it's quite obvious that
M(n) = {n} + n * M(n-1) + M(n-1)
where + means merging sets and * is an operator that adds an element to each pair or set within a set.
So M(3) = {3} + 3 * {1, 2, 12} + {1, 2, 12}
You can prove this properly inductively but idgaf, I think it's obvious enough.
now I define another function C(s) that takes a set {a, b, c...} and returns 1/q(a) + 1/q(b) + 1/q(c)...
q(x1, x2, x3...) is a function that returns x1 * x2 * x3...
fuck you, I don't know how to neatly express this in one line but it's just what you do with the numbers in the problem, right?

The thing is that C(M(n)) is f(n) (because that's how I badly defined it)
So now we get
C(M(n)) = C({n} + n * M(n-1) + M(n-1))
and I'm also not gonna bother proofing that C is kind of commutative and distributive; I don't know if those terms are right here but you can see that you can split up C over it's input set's subsets and also take out a factor easily.
f(n) = 1/n + 1/n * f(n-1) + f(n-1)
now if f(n-1) = n-1 we get that the right hand side turns out to be n
sure we need to find a start for the induction and make it formal and blah blah whatever.

This is such a clusterfuck but I swear, it's actually really simple and intuitive, I'm just too fucking stupid to write it down properly and do the inductions formally correct

>>8878859
Trivial.
Also
>He is still at a level where he has to be told which type of proof he has to construct

Topkek go back to that intro to proofs at brainlet state university.

>>8879133
oops forgot a bit that helps:

f(n) = C({n}) + C(n * M(n-1)) + C(M(n-1))
(now for that middle term, the n goes into each set within M, then after applying C the terms are multiplied, inverted and summed so I can just take out the 1/n from every term again)
f(n) = 1/n + 1/n * f(n-1) + f(n-1)

>>8879140
>>8879133
Nice.