Find the sum of the series ##\sum_{r=n+1}^{2n} u_r##

[chwala]

Homework Statement:

See attached

Relevant Equations:

Sum of series

Find question and solution here

Part (i) is clear to me as they made use of,

$$\sum_{r=n+1}^{2n} u_r=\sum_{r=1}^{2n} u_r-\sum_{r=1}^{n} u_r$$
to later give us the required working to solution...
...
$4n^2(4n+3)-n^2(2n+3)=16n^3+12n^2-2n^3-3n^2=14n^3+9n^2$ as indicated.

My question is on the second part,

I can see that they still made use of
let $r=2n, ⇒1n=r-1$ is that correct? giving us

$4n^2(4n+3)-n^2(2n+3)=r^2(2⋅r+3)-(r-1)^2(2(r-1)+3)$

$4n^2(4n+3)-n^2(2n+3)=r^2(2r+3)-(r-1)^2(2r+1)=6r^2-1$

I need more insight on the highlighted part. Thanks

 

[fresh_42]

Homework Statement:: See attached
Relevant Equations:: Sum of series

Find question and solution here

View attachment 302024

View attachment 302025

Part (i) is clear to me as they made use of,

$$\sum_{r=n+1}^{2n} u_r=\sum_{r=1}^{2n} u_r-\sum_{r=1}^{n} u_r$$
to later give us the required working to solution...
...
$4n^2(4n+3)-n^2(2n+3)=16n^3+12n^2-2n^3-3n^2=14n^3+9n^2$ as indicated.

My question is on the second part,

View attachment 302026

View attachment 302027

I can see that they still made use of
let $r=2n, ⇒1n=r-1$ is that correct? giving us

$4n^2(4n+3)-n^2(2n+3)=r^2(2⋅r+3)-(r-1)^2(2(r-1)+3)$

$4n^2(4n+3)-n^2(2n+3)=r^2(2r+3)-(r-1)^2(2r+1)=6r^2-1$

I need more insight on the highlighted part. Thanks

Why don't you use $u_r=\sum_{k=1}^r u_k - \sum_{k=1}^{r-1} u_k$?

 

[chwala]

Why don't you use $u_r=\sum_{k=1}^r u_k - \sum_{k=1}^{r-1} u_k$?

@fresh_42 let me check it out...

 

[chwala]

Why don't you use $u_r=\sum_{k=1}^r u_k - \sum_{k=1}^{r-1} u_k$?

...but isn't that what i used? let me copy paste it here,

let $r=2n, ⇒1n=r-1$ is that correct? giving us ...

 

[fresh_42]

...but isn't that what i used? let me copy paste it here,

let $r=2n, ⇒1n=r-1$ is that correct? giving us ...

I don't know. I gave up reading when I saw let $r=2n$. How could this help to find $u_r$. The reference to $2n$ and $n-1$ is misleading, to say the least.

 

[fresh_42]

Your solution is correct. Now that I did the algebra I ended up with the same result.

 

[chwala]

Your solution is correct. Now that I did the algebra I ended up with the same result.

Thanks for your input...

 

[fresh_42]

Thanks for your input...

But what was your question? Which kind of insight were you looking for?
I would write it down as
\begin{align*}
u_r&=\sum_{k=1}^r u_k-\sum_{k=1}^{r-1} u_k\\
&=r^2(2r+3)-(r-1)^2(2(r-1)+3)\\
&\ldots \\
&=6r^2-1
\end{align*}

 

[chwala]

But what was your question? Which kind of insight were you looking for?
I would write it down as
\begin{align*}
u_r&=\sum_{k=1}^r u_k-\sum_{k=1}^{r-1} u_k\\
&=r^2(2r+3)-(r-1)^2(2(r-1)+3)\\
&\ldots \\
&=6r^2-1
\end{align*}

The change of variable from $2n$ to $r$ that was the part i needed more insight...could we as well have $r+1$ and $r$? or $r+2$ and $r+1$? That is my question...

 

[fresh_42]

The change of variable from $2n$ to $r$ that was the part i needed more insight...

Yes, that was disturbing. Did post #8 answer your question?

You can also check the result with the known formulas
$\sum_{r=1}^n r^2 = \dfrac{1}{6} n (n + 1) (2 n + 1)$ and $\sum_{r=1}^n 1= n$ and see whether $\sum_{r=1}^n u_r=\sum_{r=1}^n (6n^2-1)$ yields the original formula in post #1.

 

[chwala]

Yes, that was disturbing. Did post #8 answer your question?

You can also check the result with the know formulas
$\sum_{r=1}^n r^2 = \dfrac{1}{6} n (n + 1) (2 n + 1)$ and $\sum_{r=1}^n 1= n$ and see whether $\sum_{r=1}^n u_r=\sum_{r=1}^n (6n^2-1)$ yields the original formula in post #1.

Post $8$ is clear to me...

 

[chwala]

Yes, that was disturbing. Did post #8 answer your question?

You can also check the result with the known formulas
$\sum_{r=1}^n r^2 = \dfrac{1}{6} n (n + 1) (2 n + 1)$ and $\sum_{r=1}^n 1= n$ and see whether $\sum_{r=1}^n u_r=\sum_{r=1}^n (6n^2-1)$ yields the original formula in post #1.

Yes it does, we shall have;
$6r^2=(n^2+n)(2n+1)$
$6r^2=2n^3+3n^2+n$
$6r^2-1=2n^3+3n^2$

Yap this looks more straightforward...