Understanding U_n: Solving Series Questions with Two Summations

[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

 

[micromass]

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

 

[eumyang]

Think about it.
\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.

 

[phospho]

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

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

 

[eumyang]

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

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.

 

[phospho]

Think about it.
\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.

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

 

[eumyang]

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

\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.