- #1

chwala

Gold Member

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- 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

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