All Questions
Tagged with summation algebra-precalculus
977
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Compute a tight upper bound of $\sum_{i=1}^{n-1}\frac{1}{3^i\log{n}- 3i}$?
I am trying to compute a tight upper bound of the sum below.
$\sum_{i=1}^{n-1}n\frac{\frac{1}{3^i}}{\log_3{(n/3^i)}}$
I was able to 'simplify' it up to the expression below.
$n\sum_{i=1}^{n-1}\frac{1}{...
- 185
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1
answer
86
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Can we derive the formula for summation of the series $1^{m}+2^{m}+3^{m}+...+n^{m}$ for every $m\geq2$? [duplicate]
Can we derive the formula for summation of the series $1^{m}+2^{m}+3^{m}+...+n^{m}$ for every $m\geq1$ ? For e.g. $1+2+3+...+n=\frac{n(n+1)}{2}$ . Here $m=1$ . Now I also know that $1^{2}+2^{2}+3^{2}+....
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Limit of a sum and two ways yielding two answers
$$\lim_ {n\to \infty} \sum_{r=0}^n \frac{1}{(2r+1)(2r+3)(2r+2)}$$
Now, what I did here was to first break up the general term of the sum using partial fractions, yielding
$$ \frac{1}{(2r+1)(2r+2)(2r+3)...
- 33
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How to find the sum of a finite number of increasing fractional powers
It is easy to show
$\quad\sum_{i=1}^{\infty} \dfrac{1}{n^i}=\dfrac{1}{n-1}\quad$
e.g.
$$\quad\sum_{i=1}^{\infty} \dfrac{1}{2^i}=\dfrac{1}{1}\quad$ $\quad\sum_{i=1}^{\infty} \dfrac{1}{3^i}=\dfrac{1}{2}...
- 6,416
-1
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2
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How to derive $1^3 + 2^3 + \cdots + n^3 = \left(\frac12n(n +1)\right)^2$? [duplicate]
So I know of the basic summation:
$$1 + 2 + \dots + n = \frac{n(n +1)}{2}$$
You derive this by noting that if you pair every element with the one on the other end (example: $1$ with $n$, $2$ with $n - ...
- 631
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Compute $\sum^{2024}_{k=1} \frac{2023-2022 \cos \left(\frac{\pi(2k-1)}{2024} \right)}{2021-2020 \cos \left(\frac{\pi(2k-1)}{2024} \right)}$
Question: Compute $$\sum^{2024}_{k=1} \frac{2023-2022 \cos \left(\frac{\pi(2k-1)}{2024} \right)}{2021-2020 \cos \left(\frac{\pi(2k-1)}{2024} \right)}$$
I began by rearranging the sum as follows:
$$\...
- 40.1k
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142
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Double summation problem $\sum_{i=1}^{n-1} \sum_{j=i+1}^n X_i X_j.$
I have to calculate this double summation but I am not sure I am doing it the correct way. Could you please help me with it?
The equation is: $$\sum_{i=1}^{n-1} \sum_{j=i+1}^n X_i X_j. $$
So, for ...
- 619
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How can I prove that $ \sum_{k=1}^{n}\frac{k\cdot P(n,k)}{n^{k+1}} = 1$? [duplicate]
The answer is difficult to me, I cannot figure out how to compute it.
$\sum_{k=1}^{n}\frac{k\cdot P(n,k)}{n^{k+1}}=1$
If someone can explain some technique to do it, I'd appreciate it. I tried to ...
- 1
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104
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series based on $(1+x+x^2)^n$
Question : Let $a_r$ denote the following $$(1+x+x^2)^n=\sum_{r=0}^{2n}a_rx^r$$
then prove the following
$$\sum_{r=0}^{n}(-1)^r\binom n r a_r = \begin{cases}
0 & n \ne 3k \text{ for all ...
- 3,742
1
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3
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55
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Step in proof of derivation of $1+2+\cdots+n=\tfrac{n(n+1)}{2}$.
I have been solving a few computer science problems lately and it is important for me to understand how the time complexity of an algorithm is calculated by coming up with the derivation and arriving ...
- 3,965
2
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84
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Evaluating $\sum_{k=0}^{\infty} \frac {2^k}{5^{2^k}+1}$ [duplicate]
So my teacher shared this problem with us and said everyone needs to try this, he teaches us Olympiad Math so I am assuming this wouldn't require analysis or calculus. This is the question, I have ...
- 78.1k
2
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2
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Equation with summation symbol in numerator and denominator
I am having some difficulty understanding the following formula (primarily because of presence of a summation symbol in the denominator):$$A = \frac{ \sum_{i=0}^1 (x_i+1) \times P_i} {\sum_{i=0}^1 P_i}...
- 740
3
votes
2
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122
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Check proof by induction of $\sum_{i=0}^n i^4 = \frac15 n^5+\frac12n^4+\frac13n^3-\frac1{30}n $
$\forall n \in \Bbb N$, I must demonstrate that:
$$\sum_{i=0}^n i^4 = \frac15 n^5+\frac12n^4+\frac13n^3-\frac1{30}n $$
$\bullet$ I need to prove that this is true for the first element of the sum (1):
...
- 740
31
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What is the square of summation?
Consider the following, which one of the following is true ??
$$\left( \sum^{n-1}_{j=0}Z_j\right)^2 = \sum^{n-1}_{j=0} Z_j^2 + \sum^{n-1}_{j\neq i} Z_i Z_j$$
OR
$$\left( \sum^{n-1}_{j=0}Z_j\right)^...
- 740
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Proving $\sum_{k=0}^{(m-1)/2}(-1)^k{{m+1}\choose{k}}\left(\frac{m+1}{2}-k\right)^p=0$, for odd $m\geq3$ and even $2\leqslant p\leq m-1$
I am trying to prove the following identity for odd $m\geqslant 3$ and even $2\leqslant p\leqslant m-1$:
$$\sum_{k=0}^{(m-1)/2}(-1)^k{{m+1}\choose{k}}\left(\frac{m+1}{2}-k\right)^p=0$$
When I split ...
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