All Questions
Tagged with summation discrete-mathematics
772
questions
-4
votes
1
answer
43
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Question about concrete mathematics double summation derivation [closed]
How did the author in the image convert the summation into a double summation? I can see how the double summation turns into the sum of squared integers but how would you go about converting the sum ...
-2
votes
0
answers
57
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Asymptotic notation equation problem [closed]
Hey guys this is question that we should say if both sides are equal or not and i want yall to help me find my mistake
I mentioned my question
Where is my answer’s problem?
my answer
- 15
0
votes
1
answer
61
views
What is $\sum_{m=0}^{\lfloor k/3\rfloor}{2k\choose k-3m}$ [closed]
With the floor function, I am not sure how to approach this.
Edit: I have a formula $\frac{1}{6}\left(4^k+2+3\frac{(2k)!}{(k!)^2} \right)$, but I got it purely from guess work.
- 101
2
votes
2
answers
45
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Given a set of integers, and the number of summations find the resulting frequencies
Given a set $X = \{x_1,x_2,...x_m\}\subset \mathbb{Z}$ and the number of allowed addends $N$. How can I find the frequency of every possible sum?
Example: $X = \{-1, 2\}$ and $N=3$ then every ...
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0
votes
0
answers
44
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Choosing an ordered triplet of non-negative integers $(m_1, m_2, m_3)$ such that $m_1 + m_2 + m_3 = n$.
Define
$$A = \{ (m_1, m_2, m_3) : m_1 \geq 0, m_2 \geq 0, m_3 \geq 0, m_1 + m_2 + m_3 = n\}$$
Given that $n \geq 0$ and $n, m_1, m_2, m_3 \in \mathbb{Z}^+$, then why it is the case that
$$\vert A \...
- 934
1
vote
3
answers
92
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Evaluating $\sum_{\substack{i+j+k=n \\ 0\leq i,j,k\leq n}} 1$
I need to find the sum $$\sum_{\substack{i+j+k=n \\ 0\leq i,j,k\leq n}} 1$$
For $n=1$ we have the admissible values of $(i,j,k)$ as: $(1,0,0),(0,1,0), (0,0,1)$ $$\sum_{\substack{i+j+k=1 \\ 0\leq i,j,...
- 840
1
vote
0
answers
36
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Weighted Average and System of Equations
Suppose I have all discrete distributions over the reals with support size N. So, I have a set of all possible weights $A=\{\alpha\in [0,1]^N:\sum_i^N\alpha_i=1\}$. Fix $\hat{b}$. Now suppose I have ...
- 11
0
votes
0
answers
56
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Combinatorial proof of a variation of Vandermonde's Identity
$$\sum_{r=0}^{a-b} \binom {a}{r+b} \binom {c}{d-r}=\binom{a+c}{b+d}$$
So far I have tried similar to original Vandermonde identity:
$$\sum_{k=0}^{r} \binom{a}{k} \binom {c}{r - k}= \binom{a+c}{r}$$
...
- 11
2
votes
0
answers
29
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A discrete multiple sum over strictly increasing integer sequences.
Let $l \ge 1$ and $m \ge 0$ and $k \ge l$ be integers.
Then let ${\bf A}= \left( A_\eta \right)_{\eta=1}^l$ and ${\bf B}= \left( B_\eta \right)_{\eta=1}^l$ be parameters.
We define the following ...
- 11.5k
0
votes
0
answers
33
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sum of binomial coefficient (summation formula)
i have to find the summation formula of this sum : $\sum_{k=0}^n \frac{(-1)^k}{k+1}\binom{n}k$
by using the following hint:
Develop the rational function (i.e., the quotient of two polynomials)
$$
\...
- 1
0
votes
0
answers
79
views
Sum of $2024$ variables in $[-2;3]$
Let's consider $2024$ variables $x_i$ that are arbitrary reals in $[-2;3]$ such that $x_1+x_2+...+x_{2024}=500$.
What is the greatest value of :
$$x_1^4+x_2^4+...+x_{2024}^4$$
Here is my approach but ...
- 139
1
vote
3
answers
173
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Summation with indexing variable in term [duplicate]
I am trying to solve $$\sum_{j=0}^{n-1}j2^j$$ but I don't know how to proceed with the $j$ in front of $2^j$. What is making it difficult for me is it is an indexing variable, so I don't think I can ...
- 39
0
votes
0
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38
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find number of disjoint subsets
For my discrete maths course i did the following exercise:
Let M be a finite set with n elements, find
$|\{(U,V)|U,V\subseteq M , U \not = V,U\cap V= \emptyset \}|$
I did it the following way:
choose ...
- 37
0
votes
1
answer
32
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Simplifying the inductive case for a summation
I am trying to prove by induction that $$\sum_{k=1}^n k^2(k+1) = \frac{1}{12} n(n+1)(n+2)(3n+1)$$
I proved the base case for n=1 easily, however when proving the inductive case for (n+1), I am met ...
- 39
1
vote
2
answers
55
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Explain the double sum step by step
I'm looking at the following:
But when I try to reproduce the result, I get: For $n=1$,
$$\sum_{i=0}^{n} \sum_{j=0}^{i} (i+1)(j+1)$$
$$= \sum_{i=0}^{n} (i + 1) \cdot \sum_{j=0}^{i} (j+1)$$
$$= \sum_{...
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