Series question

1. Jul 24, 2012

phospho

I don't understand how the part in yellow can give you U_n, I just don't see how taking the two summations away from each other would give U_n, could anyone explain it please

2. Jul 24, 2012

micromass

Staff Emeritus
Write it out for n=4. You'll see immediately what happens.

3. Jul 24, 2012

eumyang

$\Sigma^{n}_{r = 1} U_r= U_1 + U_2 + U_3 + ... + U_{n - 1} + U_n$, and
$\Sigma^{n - 1}_{r = 1} U_r= U_1 + U_2 + U_3 + ... + U_{n - 1}$.
So what happens when you subtract the two summations:
$\left( U_1 + U_2 + U_3 + ... + U_{n - 1} + U_n \right) - \left( U_1 + U_2 + U_3 + ... + U_{n - 1} \right)$?

EDIT: Beaten to it.

4. Jul 24, 2012

phospho

write what out? I've substituted n = 4 and get 20 if I use what they have used for part b...

5. Jul 24, 2012

eumyang

I think micromass meant this:
n = 4: $\Sigma^{4}_{r = 1} U_r= U_1 + U_2 + U_3 + U_4$
n - 1 = 3: $\Sigma^{3}_{r = 1} U_r= U_1 + U_2 + U_3$
Don't plug into the expressions with the n's.

6. Jul 24, 2012

phospho

I see, but what is the "n^2 + 4n", is that a general term or..?

7. Jul 24, 2012

eumyang

$\Sigma^{n}_{r = 1} U_r = U_1 + U_2 + U_3 + ... + U_{n - 1} + U_n = n^2 + 4n$. That was given in the problem. Notice the substitution that was made in the step after the highlighted step.