I tried to solve the following problem; but I couldn't.

Madeline Lott

Madeline Lott

Answered question

2021-12-31

I tried to solve the following problem; but I couldnt.

Answer & Explanation

Louis Page

Louis Page

Beginner2022-01-01Added 34 answers

Your answer is equivalent to the one given. In particular, we have
(4πn120+π)216=
(4πn1+20π)216=
(4π[n12+2]+20π)216=
(4π[n12]8π+20π)2}{16}=
(4π[n12]+209π)216=
porschomcl

porschomcl

Beginner2022-01-02Added 28 answers

You are correct. n1 is just an arbitrary integer.
You can rewrite,
(4πn1+209π)2=(4πn1+9π20)2=(4πn1+8π+π20)2=(4πn+π20)2
where n=n1+2
Vasquez

Vasquez

Expert2022-01-09Added 669 answers

We need
4mπ+209π=±(4πn20+π)
Considering the '-' sign,
4mπ+209π=(4πn20+π)
4m9=4n14n=4m+8n=m+2

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