1. ## Triangle 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.

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.

Hello,

to a): Plug in (n-1) instead of n into the given term:

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

to b)

Add the nth term and the (n-1)th term:

1/2n (n+1) + (1/2)n(n-1) = (1/2)n[(n+1) + (n-1)] = (1/2)n[2n] = n²

3. 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

I don't understand how you cancelled down 'a' to get 1/2n(n-1).
Hello,
Code:
(1/2)(n-1)(n    -1   +   1  ) = (1/2)(n-1) * n
\           /
equals zero

And why did you put brackets round the 1/2?

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

5. 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?

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?
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 = 1/2n² - 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)