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Anonymous Fri 02 Aug 2019 13:26:57 No.698450 View View Reply Original Report

Quoted By:

Please help me derive closed formula for 1^2 + 3^2 + 5^2 + 7^2 + ... + (2n-1)^2. I know
>closed formula for 1 + 2 + 3 + 4 + 5 + ... + n
>closed formula for 1^2 + 2^2 + 3^2 + 4^2 + ... + n^2
>closed formula for 1 + 3 + 5 + 7 + ... + 2n-1
No advanced math pls.

Anonymous

Anonymous Fri 02 Aug 2019 13:55:33 No.698452 Report

Quoted By:

>>>/sci/10841379

Anonymous

Anonymous Fri 02 Aug 2019 19:45:32 No.698554 Report

Quoted By: >>698571

https://en.wikipedia.org/wiki/Faulhaber%27s_formula#Faulhaber_polynomials

For skipping the even numbers, notice that, e.g.
(1 + 3 + 5 + 7)
=
(1 + 2 + 3 + 4 + 5 + 6 + 7)
-
(2 + 4 + 6)

And that
2 + 4 + 6
=
2(1 + 2 +3).

You should now be able to solve it.

Anonymous

Anonymous Fri 02 Aug 2019 20:53:25 No.698571 Report

Quoted By: >>698612

>>698554
>For skipping the even numbers, notice that, e.g.
>(1 + 3 + 5 + 7)
>=
>(1 + 2 + 3 + 4 + 5 + 6 + 7)
>-
>(2 + 4 + 6)
>And that
>2 + 4 + 6
>=
>2(1 + 2 +3).
I know anon, that's how I derived third formula in the OP. I really don't understand what you are aiming at anon.

Anonymous

Anonymous Sat 03 Aug 2019 00:48:16 No.698612 Report

Quoted By: >>698666

>>698571
I misread your post.
But still... you can do the same trick with squares.

1^2 + 3^2 + 5^2 + 7^2
=
1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2
-
2^2 ( 1^2 + 2^2 + 3^2 )

i.e. you use the fact that (2a)^2 = (2^2)(a^2)
also 2^2 = 4

Anonymous

Anonymous Sat 03 Aug 2019 06:01:00 No.698666 Report

Quoted By:

>>698612
Thank you anon.

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