1. ## Adding and subtracting: Rational expressions

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.

2. 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 $\displaystyle 1$ so:

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

and the second is $\displaystyle 3$, so:

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

Hence:

$\displaystyle q-p=1$

and

$\displaystyle 3q-2p=4$

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

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

RonL

3. 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.

4. 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 $\displaystyle 1(1+p)/q=1$, now assuming $\displaystyle q \ne 0$ this becomes on multiplying through by $\displaystyle q$:

$\displaystyle 1+p=q$

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

RonL