Question

Can someone explain how does step 1 render to step 2?

score 38 · Answer 1 · 2026-03-26 20:25:59

38

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Can someone explain how does step 1 render to step 2?

Published on 26 Mar 2026 - 20:25

score 39 · Answer 2 · 2019-05-03 10:30:10

39

Views

Number of Students for Equal Score

Published on 03 May 2019 - 10:30

score 391 · Answer 3 · 2026-03-26 14:22:47

391

Views

Searching for an equation (and proof of this equation) for internal path length and external path length

Published on 26 Mar 2026 - 14:22

score 1.4k · Answer 4 · 2019-05-04 20:11:25

1.4k

Views

Using induction to prove $\sum_{k=0}^{N-1} 2^k = 2^N-1$

Published on 04 May 2019 - 20:11

score 183 · Answer 5 · 2019-05-05 06:16:24

183

Views

Notation for the product of $n$ rational numbers (proof by induction)

Published on 05 May 2019 - 6:16

score 75 · Answer 6 · 2026-03-28 01:08:36

Proving $\tan^{-1}\frac{1}{2\cdot1^{2}}+\tan^{-1}\frac{1}{2\cdot2^{2}}+\cdots+\tan^{-1}\frac{1}{2\cdot n^{2}}=\frac{\pi}{4}-\tan^{-1}\frac{1}{2n+1}$

score 76 · Answer 7 · 2019-05-07 12:57:11

76

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On the inductive step in the proof by induction that $\sum_{j=1}^n(4j-1)=n(2n+1)$

Published on 07 May 2019 - 12:57

score 44 · Answer 8 · 2019-05-07 14:15:13

44

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Question using induction to prove an inequality

Published on 07 May 2019 - 14:15

score 109 · Answer 9 · 2019-05-07 17:16:29

Prove by induction: $b^n-a^n=(b-a)\cdot \sum_{k=0}^{n-1}{a^k\cdot b^{n-1-k}}$ $\forall n\in \mathbb{N}$ and $a,b\in \mathbb{R}$

score 30 · Answer 10 · 2019-05-07 20:36:25

30

Views

Base case for $\sum_{j=1}^{n-1}{j^k}<\frac{n^{k+1}}{k+1}$ with $k\in \mathbb{N}$ and $n\geq 2$

Published on 07 May 2019 - 20:36

score 68 · Answer 11 · 2026-03-26 17:22:38

68

Views

Prove that for every $n\geq6$ the equation $1/a_1^2 + ... 1/a_n^2 = 1$ has answer in $\mathbb{Z}$

Published on 26 Mar 2026 - 17:22

score 435 · Answer 12 · 2026-03-27 16:24:08

435

Views

Are there "interesting" theorems in Peano arithmetic, that only use the addition operation?

Published on 27 Mar 2026 - 16:24

score 1k · Answer 13 · 2026-03-30 05:16:43

1k

Views

n choose k proof with mathematical induction

Published on 30 Mar 2026 - 5:16

score 43 · Answer 14 · 2026-03-28 05:23:25

43

Views

For all $n\in \mathbb{N}$, find the values for$\big( \sum_{i=1}^{n} (-1)^i i! \big) \mod 36$

Published on 28 Mar 2026 - 5:23

score 62 · Answer 15 · 2019-05-09 02:25:04

62

Views

Prove that $(1+x)^n>1+nx$ for all integers $n\geq2$, where $x>-1$ and $x\neq0$.

Published on 09 May 2019 - 2:25

List Question

Can someone explain how does step 1 render to step 2?

Number of Students for Equal Score

Searching for an equation (and proof of this equation) for internal path length and external path length

Using induction to prove $\sum_{k=0}^{N-1} 2^k = 2^N-1$

Notation for the product of $n$ rational numbers (proof by induction)

Proving $\tan^{-1}\frac{1}{2\cdot1^{2}}+\tan^{-1}\frac{1}{2\cdot2^{2}}+\cdots+\tan^{-1}\frac{1}{2\cdot n^{2}}=\frac{\pi}{4}-\tan^{-1}\frac{1}{2n+1}$

On the inductive step in the proof by induction that $\sum_{j=1}^n(4j-1)=n(2n+1)$

Question using induction to prove an inequality

Prove by induction: $b^n-a^n=(b-a)\cdot \sum_{k=0}^{n-1}{a^k\cdot b^{n-1-k}}$ $\forall n\in \mathbb{N}$ and $a,b\in \mathbb{R}$

Base case for $\sum_{j=1}^{n-1}{j^k}<\frac{n^{k+1}}{k+1}$ with $k\in \mathbb{N}$ and $n\geq 2$

Prove that for every $n\geq6$ the equation $1/a_1^2 + ... 1/a_n^2 = 1$ has answer in $\mathbb{Z}$

Are there "interesting" theorems in Peano arithmetic, that only use the addition operation?

n choose k proof with mathematical induction

For all $n\in \mathbb{N}$, find the values for$\big( \sum_{i=1}^{n} (-1)^i i! \big) \mod 36$

Prove that $(1+x)^n>1+nx$ for all integers $n\geq2$, where $x>-1$ and $x\neq0$.

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