Question

From $x^4+4x^3+x^2-6x-1=0$ then find $S=\frac{m}{n}=\sum_{i=1}^{4}\frac{1}{\alpha_i^4-15\alpha_i^2-10\alpha_i+24}$

score 112 · Answer 1 · 2026-04-17 10:35:44

112

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From $x^4+4x^3+x^2-6x-1=0$ then find $S=\frac{m}{n}=\sum_{i=1}^{4}\frac{1}{\alpha_i^4-15\alpha_i^2-10\alpha_i+24}$

Published on 17 Apr 2026 - 10:35

score 67 · Answer 2 · 2026-04-25 08:00:32

67

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Closed form for a multilinear skew-symmetric map applied in sums?

Published on 25 Apr 2026 - 8:00

score 34 · Answer 3 · 2026-04-16 04:29:33

34

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Any hope for evaluating, or accurately approximating, this sum for large $n$?

Published on 16 Apr 2026 - 4:29

score 271 · Answer 4 · 2026-04-13 02:34:19

271

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Closed-form summation of exponential

Published on 13 Apr 2026 - 2:34

score 48 · Answer 5 · 2026-04-15 15:21:13

48

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How do I evaluate $\sum_{i=0}^{n-1}2^i+1$ and $\sum_{i=2}^{n-1} \log i^{n}$?

Published on 15 Apr 2026 - 15:21

score 191 · Answer 6 · 2026-04-12 04:02:09

191

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On the conjecture that $1-\frac 13 + \frac 16 +\frac 1{10} -\frac 1{15}+\cdots = 1\frac 19$

Published on 12 Apr 2026 - 4:02

score 2.2k · Answer 7 · 2026-03-25 11:03:28

2.2k

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On proving that $\sum\limits_{n=1}^\infty \frac{n^{13}}{e^{2\pi n}-1}=\frac 1{24}$

Published on 25 Mar 2026 - 11:03

score 73 · Answer 8 · 2026-04-15 21:15:06

73

Views

Prove that this sum of products of binomial coefficient can be simplified like this

Published on 15 Apr 2026 - 21:15

score 37 · Answer 9 · 2026-04-12 13:45:08

37

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Compute the value of 2 summation equations

Published on 12 Apr 2026 - 13:45

score 1.2k · Answer 10 · 2026-04-14 09:46:16

1.2k

Views

Rewrite of sum into parts

Published on 14 Apr 2026 - 9:46

score 887 · Answer 11 · 2026-04-13 14:32:41

887

Views

If $f(r)=r(r+1)(r+2)(r+3)$, simplify $f(r+1)-f(r)$ and use your result to find

Published on 13 Apr 2026 - 14:32

score 94 · Answer 12 · 2026-04-16 01:13:15

94

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Prove $\binom{n}{0} + \frac{1}{2} \binom{n}{1} + \frac{1}{3} \binom{n}{2} + ... + \frac{1}{n+1} \binom{n}{n} = \frac{2^{n + 1} - 1}{n + 1}$.

Published on 16 Apr 2026 - 1:13

score 74 · Answer 13 · 2026-04-10 03:38:07

74

Views

Proof of a summation identity

Published on 10 Apr 2026 - 3:38

score 82 · Answer 14 · 2026-04-16 15:18:25

82

Views

What is the value of $[(1^ {-2/3}) + (2^ {-2/3}) + (3^ {-2/3}) + … + (1000^ {-2/3})] $? Where $[x]$ stands for the greatest integer function.

Published on 16 Apr 2026 - 15:18

score 172 · Answer 15 · 2026-04-09 02:09:30

172

Views

Egyptian fractions

Published on 09 Apr 2026 - 2:09

List Question

From $x^4+4x^3+x^2-6x-1=0$ then find $S=\frac{m}{n}=\sum_{i=1}^{4}\frac{1}{\alpha_i^4-15\alpha_i^2-10\alpha_i+24}$

Closed form for a multilinear skew-symmetric map applied in sums?

Any hope for evaluating, or accurately approximating, this sum for large $n$?

Closed-form summation of exponential

How do I evaluate $\sum_{i=0}^{n-1}2^i+1$ and $\sum_{i=2}^{n-1} \log i^{n}$?

On the conjecture that $1-\frac 13 + \frac 16 +\frac 1{10} -\frac 1{15}+\cdots = 1\frac 19$

On proving that $\sum\limits_{n=1}^\infty \frac{n^{13}}{e^{2\pi n}-1}=\frac 1{24}$

Prove that this sum of products of binomial coefficient can be simplified like this

Compute the value of 2 summation equations

Rewrite of sum into parts

If $f(r)=r(r+1)(r+2)(r+3)$, simplify $f(r+1)-f(r)$ and use your result to find

Prove $\binom{n}{0} + \frac{1}{2} \binom{n}{1} + \frac{1}{3} \binom{n}{2} + ... + \frac{1}{n+1} \binom{n}{n} = \frac{2^{n + 1} - 1}{n + 1}$.

Proof of a summation identity

What is the value of $[(1^ {-2/3}) + (2^ {-2/3}) + (3^ {-2/3}) + … + (1000^ {-2/3})] $? Where $[x]$ stands for the greatest integer function.

Egyptian fractions

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