+0  
 
0
37
9
avatar+438 

Compute the sum of

\((a +(2n+1)d)^2- (a + (2n)d)^2 +(a + (2n-1)d)^2 - (a+(2n-2)d)^2 + \cdots + (a+d)^2 - a^2\)

 

 

thank you to anyone who helps!

:)))

 May 22, 2020
 #1
avatar+675 
+3

Hi Loki, I feel like this problem is very confusing to me, sorry:(

 
 May 22, 2020
 #2
avatar+438 
+1

yeah it's super confusing to me too :(((

 
lokiisnotdead  May 22, 2020
 #3
avatar+1498 
+3

So we have this where we have

 

(a * (x)) ^2.

 

x ranges from 2n+ 1 to 1, so there are thus 2n terms in this sequence.

 

(a + 2nd+d) ^ 2= a^2 + 2and + ad + 4n^2d^2 + 2and + 2nd^2 + d^2 + 2nd^2 + ad.

 

Simplifying that is a^2 + (2nd) ^2 + d^2 + 2ad + 4and + 4(nd)^2 + 2n(d)^2.

 

The next term is 
 

(a + 2nd)^2 = a^2 + 4and + 4(nd)^2

 

So we subtract and get... 

a^2 + (2nd) ^2 + d^2 + 2ad + 4and + 4(nd)^2 + 2n(d)^2

                           d^2 + 2ad                              + 2n(d)^2??
 

Yeah...

 

Imma use a expander caculator.

 

-2ad+4adn+4d^2n^2-4d^2n+d^2+a^2

 

and the next term...

 

-4ad+4adn+4d^2n^2-8d^2n+4d^2+a^2

 

So we have now...

 

-2ad+4adn+4d^2n^2-4d^2n+d^2+a^2

-4ad+4adn+4d^2n^2-8d^2n+4d^2+a^2

 

That's -2ad -4d^2n + 3d^2 

 

Last time was

 

2ad + 2d^2n +2ad

-2ad -4d^2n + 3d^2 

 

So minus -4ad ...

 

I give up

 

sorry

 
 May 22, 2020
edited by hugomimihu  May 22, 2020
 #4
avatar
-1

This is an AoPS homework problem.  Please do not post solutions.

 
 May 22, 2020
 #5
avatar+675 
+4

are you sure, 'cause usually hugo would have said that first

 
 May 22, 2020
 #7
avatar+1498 
+2

Possibly. I self learn algebra and I don't think this is in a textbook, but I don't take AoPS courses on algebra so idk.

 

Besides I gave a partial answer that idk is correct or not.

 
hugomimihu  May 22, 2020
edited by hugomimihu  May 22, 2020
 #6
avatar+675 
+3

hehehe:)

 
 May 22, 2020
 #8
avatar+1498 
+3

Let me try again. 

 

Difference of first pair is

 

2ad+d^2+4dn

 

Second pair

2ad-3d^2+4dn

 

So we seem to subtract 4d^2 every iteration.

 

For the third pair it's 

2ad-7d^2+4dn

Great!

 

As I said there are 2n terms in this series, so we have n pairs.

 

Thus we have 2adn + 4d(n^2), but also...

 

The x(d^2). So thus we have (1 - 3- 7- 11 - 15... - (1-4n))(d^2.)

 

That's a start.

 

Loki I don't know how to solve the question but perhaps this may help you , I don't know.

 

I was using

 

https://www.symbolab.com/solver/expand-calculator/expand%20

 

This for expansions.

 
 May 22, 2020
edited by hugomimihu  May 22, 2020
 #9
avatar+30028 
+1

This question has appeared several times previously.  Search the answers of heureka or myself.

 
 May 22, 2020

4 Online Users