Triangle number sequence: 1,3,6,10,15,21,....

nth term is given by 1/2n (n+1)

a. give algebraic expression for the (n-1) term of the sequence.

b. Prove that the sum of any two consecutive triangle numbers is a square number.

Printable View

- May 18th 2007, 01:05 AMGAdamsTriangle Numbers
Triangle number sequence: 1,3,6,10,15,21,....

nth term is given by 1/2n (n+1)

a. give algebraic expression for the (n-1) term of the sequence.

b. Prove that the sum of any two consecutive triangle numbers is a square number. - May 18th 2007, 02:34 AMearboth
- May 18th 2007, 05:36 AMGAdams
I don't understand how you cancelled down 'a' to get 1/2n(n-1). And why did you put brackets round the 1/2?

Thanks - May 18th 2007, 07:38 AMearboth
Hello,

Code:`(1/2)(n-1)(n -1 + 1 ) = (1/2)(n-1) * n`

\ /

equals zero

As long as Latex isn't working writing mathematical expressions is a little bit tricky. Only to exclude any abiguity I use brackets to write a fraction. If you writeCode:`1`

1/2n(n-1) sometimes it is meant to be: --------

2n(n-1)

- May 18th 2007, 08:51 AMGAdams
Hmmm.

I have it up to here:

(1/2) (n-1) (n-1 + 1)

You cancelled the -1 and +1 in the last brackets to get:

(1/2) (n-1) *n

Which to my 'logical' mind would multiply out to give:

1/2n - 1/2 *n

Which is zero! It seems that you have multiplied the 1/2 by the last n and left the brackets intact.

How would I know to do that? - May 18th 2007, 11:30 AMearboth
Hi,

there are 2 points where I was obviously not exact enough:

(1/2) (n-1) (n-1 + 1) = (1/2) (n-1) (n**+ 0**) = (1/2) (n-1) (n)

(1/2) (n-1) (n-1 + 1) = (1/2) (n-1) *n =**1/2nē - 1/2 *n**. Because you multiply a sum you must use the law of distribution (not certain if this is the appropriate expression in English)