+738 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!
:)))
+1005 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
+1005 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.
+1005 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.
