# Math Help - Basic proof.

1. ## Basic proof.

Proof that the rth term can be found by finding the difference between the sum of the first r terms and the sum of the first (r-1) terms: That is, $U_{r} = S_{r}-S_{r-1}$.

What I really don't understand is $S_{r-1}$. For example, what is the $S_{r-1}$ of $3+5+7+9$? Is it $0+3+5+7$? Or $1+3+5+7$?

2. $S_{r}$ is the sum of terms up to and including $U_{r}$
$S_{r-1}$ is the sum of terms up to and including $U_{r-1}$

i.e.
$S_{r}=U_{1}+U_{2}+U_{3}+...+U_{r-1}+U_{r}$ -----(1)
$S_{r-1}=U_{1}+U_{2}+U_{3}+...+U_{r-2}+U_{r-1}$ -----(2)

(1)-(2): $U_{r}=S_{r}-S_{r-1}$

3. Originally Posted by acc100jt
$S_{r}$ is the sum of terms up to and including $U_{r}$
$S_{r-1}$ is the sum of terms up to and including $U_{r-1}$

i.e.
$S_{r}=U_{1}+U_{2}+U_{3}+...+U_{r-1}+U_{r}$ -----(1)
$S_{r-1}=U_{1}+U_{2}+U_{3}+...+U_{r-2}+U_{r-1}$ -----(2)

(1)-(2): $U_{r}=S_{r}-S_{r-1}$
Yeah. I see. But how on Earth would you come up with that first? I mean, say you wanted to come up with this formula for the rth term, and you don't know that it's $S_{r}-S_{r-1}$? What we did now there was to confirm that $U_{r} = S_{r}-S_{r-1}$. There is no way I could have simply thought: Ah, find the difference between the sum of the first r terms and the sum of the first (r-1) terms.

4. Look at how $S_{r}$ is defined.

$
S_{r}=U_{1}+U_{2}+U_{3}+...+U_{r-1}+U_{r}
$

Then consider
$
S_{r-1}=U_{1}+U_{2}+U_{3}+...+U_{r-2}+U_{r-1}
$

If I take the difference, then I get a formula for the rth term, $U_{r}=S_{r}-S_{r-1}$ right?

5. Originally Posted by acc100jt
Look at how $S_{r}$ is defined.

$
S_{r}=U_{1}+U_{2}+U_{3}+...+U_{r-1}+U_{r}
$

Then consider
$
S_{r-1}=U_{1}+U_{2}+U_{3}+...+U_{r-2}+U_{r-1}
$

If I take the difference, then I get a formula for the rth term, $U_{r}=S_{r}-S_{r-1}$ right?
Oh, right.

I didn't see the $U_{r}$ at the end of the $S_{r}$ defintion.

Many thanks, Acc100jt.