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## #1 2018-02-02 14:56:34

Monox D. I-Fly
Member
From: Indonesia
Registered: 2015-12-02
Posts: 1,047

Determine the value of n if 1 + 3 + 6 + ... +

n(n - 1) = 364

What I did:
I know that 1, 3, and 6 are the result of arithmetic series with the starting value 1 and the difference 2, thus that sum can be written as

+
+
+ ... +
= 364. However, by assuming that
=
n(n - 1) I got n + 1 = n - 1 which is simply unsolvable at all. After all, the term
n(n - 1) doesn't match for n = 1. Does this question even have any solution?

Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away. May his adventurous soul rest in peace at heaven.

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## #2 2018-02-02 20:03:36

zetafunc
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Registered: 2014-05-21
Posts: 2,179
Website

Monox D. I-Fly wrote:

I know that 1, 3, and 6 are the result of arithmetic series with the starting value 1 and the difference 2, thus that sum can be written as

+
+
+ ... +
= 364.

No: this is not an arithmetic series. The terms form the sequence of triangular numbers, whose nth term is
(which you can also write as
, though I prefer the former).

You have been asked to find the value of
such that
That last term is the nth term of the sequence
. Notice that if
, then
. Similarly taking
gives you
, and so on. So what you actually want to do is find the value of
such that:

Do you know how to do that?

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## #3 2018-02-02 20:10:08

Monox D. I-Fly
Member
From: Indonesia
Registered: 2015-12-02
Posts: 1,047

Other than

, no.

Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away. May his adventurous soul rest in peace at heaven.

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## #4 2018-02-02 20:27:52

zetafunc
Moderator
Registered: 2014-05-21
Posts: 2,179
Website
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