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

• chwala
In summary: ThanksIn summary, the conversation discusses the use of the formula $$\sum_{r=n+1}^{2n} u_r=\sum_{r=1}^{2n} u_r-\sum_{r=1}^{n} u_r$$ to solve a problem. The question is raised about the use of a variable substitution, specifically, whether using ##r=2n## is correct. The discussion concludes that this substitution is indeed valid and yields the desired result.
chwala
Gold Member
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

chwala said:
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##?

fresh_42 said:
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...

fresh_42 said:
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 ...

chwala said:
...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.

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

fresh_42 said:
Your solution is correct. Now that I did the algebra I ended up with the same result.

chwala said:
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*}

fresh_42 said:
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...

chwala said:
The change of variable from ##2n## to ##r## that was the part i needed more insight...

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.

fresh_42 said:

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...

fresh_42 said:

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...

fresh_42

## What is the formula for finding the sum of a series?

The formula for finding the sum of a series is:
Sum = (n/2)(a + l)
Where n is the number of terms in the series, a is the first term, and l is the last term.

## What is the difference between an arithmetic and geometric series?

An arithmetic series is a series where each term is obtained by adding a constant value to the previous term. A geometric series is a series where each term is obtained by multiplying the previous term by a constant value. The formula for finding the sum of an arithmetic series is different from that of a geometric series.

## What is the significance of the index in a series?

The index in a series represents the position of a term in the series. In the given series, the index ranges from n+1 to 2n, meaning there are n terms in the series. The index helps in determining the number of terms and the starting and ending points of the series.

## How can I simplify the given series to find the sum?

To simplify the given series, you can use the formula for finding the sum of an arithmetic series:
Sum = (n/2)(a + l)
In this case, n = n, a = un+1, and l = u2n. You can then plug in the values and simplify to find the sum.

## What are the applications of finding the sum of a series?

Finding the sum of a series is useful in various mathematical and scientific fields. It is used in calculus to calculate definite integrals, in statistics to find the mean and standard deviation, and in finance to calculate compound interest. It is also used in physics and engineering to calculate the total energy or work done in a system.

• Calculus and Beyond Homework Help
Replies
3
Views
408
• Calculus and Beyond Homework Help
Replies
2
Views
728
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
247
• Calculus and Beyond Homework Help
Replies
4
Views
298
• Calculus and Beyond Homework Help
Replies
14
Views
1K
• Calculus and Beyond Homework Help
Replies
17
Views
603
• Calculus and Beyond Homework Help
Replies
20
Views
1K
• Calculus and Beyond Homework Help
Replies
7
Views
2K
• Calculus and Beyond Homework Help
Replies
9
Views
1K