All Questions
Tagged with summation algebra-precalculus
165
questions
7
votes
3
answers
12k
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Sum of first n natural numbers proof
I know how to prove this by induction but the text I'm following shows another way to prove it and I guess this way is used again in the future. I'm confused by it.
So the expression for first n ...
- 639
7
votes
3
answers
317
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Are there $a,b \in \mathbb{N}$ that ${(\sum_{k=1}^n k)}^a = \sum_{k=1}^n k^b $ beside $2,3$
We know that:
$$\left(\sum_{k=1}^n k\right)^2 = \sum_{k=1}^n k^3 $$
My question is there other examples that satisfies:
$$\left(\sum_{k=1}^n k\right)^a = \sum_{k=1}^n k^b $$
- 1,563
6
votes
4
answers
3k
views
if 1, $\alpha_1$, $\alpha_2$, $\alpha_3$, $\ldots$, $\alpha_{n-1}$ are nth roots of unity then...
if 1, $\alpha_1$, $\alpha_2$, $\alpha_3$, $\ldots$, $\alpha_{n-1}$ are nth roots of unity then
$$\frac{1}{1-\alpha_1} + \frac{1}{1-\alpha_2} + \frac{1}{1-\alpha_3}+\ldots+\frac{1}{1-\alpha_n} = ?$$
...
- 1,396
6
votes
3
answers
1k
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Explain why calculating this series could cause paradox?
$$\ln2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots
= (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) - 2(\frac{1}{2} + \frac{1}{4} + \cdots)$$
$$= (1 + \frac{1}{2} + \frac{...
- 8,038
3
votes
1
answer
1k
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Please help explain how "Any symmetric sum can be written as a Polynomial of the elementary symmetric sum functions"
I was browsing through the Art Of Problem Solving website and came across this:
"Any symmetric sum can be written as a Polynomial of the elementary symmetric sum functions, for example
$x^3 + y^3 + ...
- 637
3
votes
6
answers
194
views
Sum the series $\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$
$\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$
The general term seems to be
$$T_r= \frac{r+2}{r(r+1)(r+3)}.$$
I see no way to telescope this because the factors of the ...
- 2,132
3
votes
2
answers
9k
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Proof of a summation of $k^4$
I am trying to prove $$\sum_{k=1}^n k^4$$
I am supposed to use the method where $$(n+1)^5 = \sum_{k=1}^n(k+1)^5 - \sum_{k=1}^nk^5$$
So I have done that and and after reindexing and a little algebra, ...
- 419
3
votes
4
answers
174
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Evaluate $\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}$, where $q^7=1$ and $q\neq 1$.
Let $q$ be a complex number such that $q^7=1$ and $q\neq 1$. Evaluate $$\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}.$$
The given answer is $\frac{3}{2}$ or $-2$. But my answer is $\pm 2$.
...
- 373
2
votes
2
answers
852
views
Find the $\frac mn$ if $T=\sin 5°+\sin10°+\sin 15°+\cdots+\sin175°=\tan \frac mn$
It's really embarrassing to be able to doesn't solve this simple-looking trigonometry question.
$$T=\sin(5^\circ) +\sin(10^\circ) + \sin(15^\circ) + \cdots +\sin(175^\circ) =\tan \frac mn$$
Find the ...
user548054
2
votes
2
answers
391
views
Problem with Abel summation
Let $$S_n=\sum_{i=1}^{n}\sin k,\quad S_0=0.$$
Then
$$\sum_{k=1}^{n}\frac{\sin k}{k}=\sum_{k=1}^{n}\frac{S_k-S_{k-1}}{k}=\frac{S_n}{n}+\sum_{k=1}^{n-1}\frac{S_k}{k(k+1)}.$$
Could someone please ...
- 3,017
2
votes
4
answers
258
views
The sum of powers of $2$ between $2^0$ and $2^n$
Lately, I was wondering if there exists a closed expression for $2^0+2^1+\cdots+2^n$ for any $n$?
- 217
1
vote
2
answers
98
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If $S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\cdots+\frac{n}{1+n^2+n^4}$, then calculate $14S$.
If $$S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\cdots+\frac{n}{1+n^2+n^4}\,$$ find the value of $14S$.
The question can be simplified to:
Find $S=\sum\limits_{k=1}^n\,t_k$ if $t_n=\dfrac{n}{1+n^2+n^...
- 2,642
-2
votes
2
answers
306
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Prove summation inequality with induction
Yesterday at university a professor gave us two problems that left many doubts.
1) $\displaystyle \sum_{i=1}^n \frac{1}{i^2} \leq 2-\frac1{n}$,
2) $\displaystyle \sum_{i=1}^n \frac1{n+i} \leq \...
- 327
39
votes
5
answers
76k
views
Simple Double Summation
I've seen how nesting works with a simple $(i+j)$ but this problem below is tripping me up. It's either because of the multipliers or because they each start at zero but I get 60, and the answer I ...
- 661
29
votes
2
answers
829
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How to prove $\sum_{n=0}^{\infty} \frac{1}{1+n^2} = \frac{\pi+1}{2}+\frac{\pi}{e^{2\pi}-1}$
How can we prove the following
$$\sum_{n=0}^{\infty} \dfrac{1}{1+n^2} = \dfrac{\pi+1}{2}+\dfrac{\pi}{e^{2\pi}-1}$$
I tried using partial fraction and the famous result $$\sum_{n=0}^{\infty} \...
user208998