Math Is Fun Forum

Monox D. I-Fly

Member

From: Indonesia

Registered: 2015-12-02

Posts: 2,000

[ASK] Induction Again

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?

---

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zetafunc

Moderator

Registered: 2014-05-21

Posts: 2,440

Website

Re: [ASK] Induction Again

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?

Monox D. I-Fly

Member

From: Indonesia

Registered: 2015-12-02

Posts: 2,000

Re: [ASK] Induction Again

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.

zetafunc

Moderator

Registered: 2014-05-21

Posts: 2,440

Website

Re: [ASK] Induction Again

Have you come across these formulae before?

If not, then one approach (which fits the title of the thread more accurately, I suppose) is to prove this result by induction:

and then once you have, set that equal to 364 and solve.