Show that $1+2+...+(n-1)\equiv 0 \mod n$

Question

Show that if $n$ is a positive integer, then

$1+2+...+(n-1)\equiv 0 \mod n$

My attempt:

If $n=1$, $0\equiv 0 \mod 1$

Assume that, $1+2+...+(k-1)\equiv 0 \mod k$ for some $k\in \Bbb{N}$.

$1+2+...+(k-1)+k\equiv (0+k) \mod k $

$\equiv k \mod k$

$\equiv 0 \mod k$

But I want to show that $1+2+...+(k-1)+k\equiv 0 \mod (k+1) $, how can I Do it please?

@BillDubuque This question is much better written than the one you have linked - that one shows no effort and is badly formatted. — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Feb 23 '19 at 19:12
@TheSimpliFire The pedagogical value lies in the answers, not the question. This question has been asked and answered a countless number of times. — Bill Dubuque, Feb 23 '19 at 19:14
@BillDubuque Answers usually do not influence the quality of the question. But the duplicate you have now set is fine. — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Feb 23 '19 at 19:17

score 1 · Accepted Answer · answered Feb 23 '19 at 19:03

1

The statement is false for even $n$. For odd $n$, note that $\sum_{i=1}^{n-1}i=\frac{n(n-1)}2$.

answered Feb 23 '19 at 19:03

ə̷̶̸͇̘̜́̍͗̂̄︣͟

27,005

Thank you so much. $1+2+...+(n-1)=\frac{n(n-1)}{2}=(\frac{n-1}{2}) n \equiv 0 \mod n$ ? – Dima Feb 23 '19 at 19:10
Yes, provided $n$ is odd. Otherwise, $n-1$ will not be divisible by $2$. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Feb 23 '19 at 19:11
You are right, Thanks. – Dima Feb 23 '19 at 19:13

score 0 · Answer 2 · answered Feb 23 '19 at 19:03

0

Hint: Use that $$\sum_{i=1}^{n-1}i=\frac{1}{2}n(n-1)$$

answered Feb 23 '19 at 19:03

Dr. Sonnhard Graubner

95,283

score 0 · Answer 3 · answered Feb 23 '19 at 19:10

0

If you don't want to use the summation formula for the first $n$ integers, you can use that for every $k\in \mathbb{Z}_n$ there exists a unique $k'\in \mathbb{Z}_n$ such that $k + k' = 0$.

answered Feb 23 '19 at 19:10

Count Iblis

10,366

Show that $1+2+...+(n-1)\equiv 0 \mod n$

3 Answers3