+0

0
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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
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Hi Loki, I feel like this problem is very confusing to me, sorry:(

May 22, 2020
#2
+732
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yeah it's super confusing to me too :(((

#3
+976
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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

Yeah...

Imma use a expander caculator.

and the next term...

So we have now...

Last time was

I give up

sorry

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

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

May 22, 2020
#5
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are you sure, 'cause usually hugo would have said that first

May 22, 2020
#7
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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
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hehehe:)

May 22, 2020
#8
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Let me try again.

Difference of first pair is

Second pair

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

For the third pair it's

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
+30929
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This question has appeared several times previously.  Search the answers of heureka or myself.

May 22, 2020