Question

Find $k>0$ such that $(k+1)+\sum^{\infty}_{n=1} \left(\frac{\binom{4n}{n}}{3n+1} k^{2n}+\frac{2\binom{4n+1}{n}}{3n+2} k^{2n+1} \right)=2$

score 88 · Answer 1 · 2026-04-16 17:24:42

88

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Find $k>0$ such that $(k+1)+\sum^{\infty}_{n=1} \left(\frac{\binom{4n}{n}}{3n+1} k^{2n}+\frac{2\binom{4n+1}{n}}{3n+2} k^{2n+1} \right)=2$

Published on 16 Apr 2026 - 17:24

score 55 · Answer 2 · 2026-04-09 10:46:22

55

Views

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score 60 · Answer 3 · 2026-04-13 10:18:19

60

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score 131 · Answer 4 · 2026-04-14 20:47:44

131

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score 56 · Answer 5 · 2026-04-15 16:53:19

56

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score 25 · Answer 6 · 2026-04-14 18:32:13

25

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score 62 · Answer 7 · 2026-04-15 21:17:30

62

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score 121 · Answer 8 · 2026-04-09 18:25:58

121

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score 55 · Answer 9 · 2026-04-12 10:10:54

55

Views

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score 73 · Answer 10 · 2026-04-14 00:31:59

73

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score 57 · Answer 11 · 2026-04-08 21:43:56

57

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score 97 · Answer 12 · 2026-04-14 21:51:02

97

Views

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Published on 14 Apr 2026 - 21:51

score 345 · Answer 13 · 2026-04-13 17:11:38

345

Views

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Published on 13 Apr 2026 - 17:11

score 117 · Answer 14 · 2026-04-08 22:27:18

117

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score 57 · Answer 15 · 2026-04-17 13:57:54

57

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Published on 17 Apr 2026 - 13:57

List Question

Find $k>0$ such that $(k+1)+\sum^{\infty}_{n=1} \left(\frac{\binom{4n}{n}}{3n+1} k^{2n}+\frac{2\binom{4n+1}{n}}{3n+2} k^{2n+1} \right)=2$

Different representations of a generating function

Pascals triangle and binomial theory $(1+x)^n$ expansion

how to prove $\sum_{j=\frac{k}{2}+1}^{k-1}\binom{k-1}{j}\binom{n-(k-1)}{k-j}<2\binom{n-1}{k-1}$?

Searching for a reference on a congruence relation

Relation between one double recursive formula and numbers q-binomial

Combinatorial interpretation of the number of subsequences of fixed length that preserve relative order

Upper bound on sum of squares of binomial probabilities

Name for variant of Pascal's triangle with different initial conditions

Calculating the limit of a sequence that includes Binomial Coefficients.

An attempt for finding the number of integer solutions with restrictions

For what $a \in N$ , Is $\sum_{k=0} ^ n (-1)^k k^a \binom nk = 0 ?$

Evaluation of $\sum^{19}_{k=1}\frac{(-1)^{k-1}k}{\binom{20}{k}}$

An approximation to a binomial looking formula

Simple integer sequence with unknown direct generator

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