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!

:)))

lokiisnotdead May 22, 2020

#3**+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

hugomimihu May 22, 2020

#5

#7**+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

#8**+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.

hugomimihu May 22, 2020