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.
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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.
Quote:
Originally Posted by GAdams
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] = nē
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
Hello,Quote:
Originally Posted by GAdams
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 writeQuote:
Originally Posted by GAdams
Code:1
1/2n(n-1) sometimes it is meant to be: --------
2n(n-1)
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,Quote:
Originally Posted by GAdams
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)