# Adding and subtracting: Rational expressions

• Feb 20th 2008, 01:32 PM
mike_302
Normally I wouldn't post the question straight out of the book, but today I am desperate because I just cannot get any bearing on this one.

Triangular numbers of objects can be arranged to form triangles. The first four triangle numbers are: 1, 3, 6, 10.
a)an expression to find the nth triangular number can be written in the form n(n+p)/q where p and q represent whole numbers. Complete the expression by determining the numbers represented by p and q.
• Feb 20th 2008, 02:08 PM
CaptainBlack
Quote:

Originally Posted by mike_302
Normally I wouldn't post the question straight out of the book, but today I am desperate because I just cannot get any bearing on this one.

Triangular numbers of objects can be arranged to form triangles. The first four triangle numbers are: 1, 3, 6, 10.
a)an expression to find the nth triangular number can be written in the form n(n+p)/q where p and q represent whole numbers. Complete the expression by determining the numbers represented by p and q.

Well, the first triangular number is $1$ so:

$1(1+p)/q=1$

and the second is $3$, so:

$2(2+p)/q=3$

Hence:

$q-p=1$

and

$3q-2p=4$

so $q=2$, and $p=1$.

Hence $T_n=\frac{n(n+1)}{2}$

RonL
• Feb 20th 2008, 02:14 PM
mike_302
Sorry, It was explained simply, and I like that, but you lost me at:

Hence: q-p=1 ... I do not understand how you obtained that.
• Feb 20th 2008, 02:24 PM
CaptainBlack
Quote:

Originally Posted by mike_302
Sorry, It was explained simply, and I like that, but you lost me at:

Hence: q-p=1 ... I do not understand how you obtained that.

The first equation I had was $1(1+p)/q=1$, now assuming $q \ne 0$ this becomes on multiplying through by $q$:

$1+p=q$

which is then rearranged to give $q-p=1$

RonL