Question

Writing $a^2b^0c^0+a^0b^2c^0+a^0b^0c^2+a^1b^1c^0+a^1b^0c^1+a^0b^1c^1$ using $\sum$?

score 157 · Answer 1 · 2026-04-04 01:54:47

157

Views

Writing $a^2b^0c^0+a^0b^2c^0+a^0b^0c^2+a^1b^1c^0+a^1b^0c^1+a^0b^1c^1$ using $\sum$?

Published on 04 Apr 2026 - 1:54

score 185 · Answer 2 · 2019-06-30 22:32:46

185

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Showing $\left\lfloor \sum_{k=1}^{10000} {1 \over \sqrt k}\right\rfloor = 198$

Published on 30 Jun 2019 - 22:32

score 70 · Answer 3 · 2026-03-25 03:07:26

70

Views

Simplifcation of an summation of gamma functions by Mathematica

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score 61 · Answer 4 · 2019-07-02 10:19:00

61

Views

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120

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score 29 · Answer 6 · 2019-07-03 00:04:33

29

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score 954 · Answer 7 · 2026-03-26 09:42:40

954

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118

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score 173 · Answer 9 · 2019-07-03 17:44:19

173

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score 48 · Answer 10 · 2019-07-03 19:31:32

48

Views

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score 140 · Answer 11 · 2026-03-26 06:33:44

140

Views

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score 2.3k · Answer 12 · 2026-04-05 20:54:16

2.3k

Views

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Published on 05 Apr 2026 - 20:54

score 99 · Answer 13 · 2019-07-04 15:31:02

99

Views

Prove that $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\ln(2n+1)}{2n+1} = \frac{\pi}{4}(\gamma - \ln \pi) + \pi \ln \Gamma(\frac{3}{4})$

Published on 04 Jul 2019 - 15:31

score 96 · Answer 14 · 2026-03-26 19:18:48

96

Views

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Published on 26 Mar 2026 - 19:18

score 49 · Answer 15 · 2026-03-25 22:03:58

49

Views

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Published on 25 Mar 2026 - 22:03

List Question

Writing $a^2b^0c^0+a^0b^2c^0+a^0b^0c^2+a^1b^1c^0+a^1b^0c^1+a^0b^1c^1$ using $\sum$?

Showing $\left\lfloor \sum_{k=1}^{10000} {1 \over \sqrt k}\right\rfloor = 198$

Simplifcation of an summation of gamma functions by Mathematica

Help with calculation and intuition for the sums $\sum_{i=1}^{n}1$ and $\sum_{i=0}^{n-1}1$

Difficult limit of an infinite sum

Is a One-Time Pyment Better Than a Saving Plan (Stocks, Funds)?

Sum of all the numbers in the grid.

Function that grows as fast as sum of previous values

How to find 50-th term of this sequence and its sum

Why can the error term in the sum definition of an integral be reduced to O(h)?

Summation containing $\pi$ gives an expression containing $e$

Sum of number's digits

Prove that $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\ln(2n+1)}{2n+1} = \frac{\pi}{4}(\gamma - \ln \pi) + \pi \ln \Gamma(\frac{3}{4})$

Concrete mathematics $\sum_{0 \le k \lt n} (\sqrt {\lfloor k \rfloor})$

Better bound for an equation of sums

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