I need to compute: \(\displaystyle{\tan{{\left({\arctan{{\left({\frac{{{1}}}{{{2}}}}\right)}}}+{\arctan{{\left({\frac{{{2}}}{{{9}}}}\right)}}}+{\arctan{{\left({\frac{{{1}}}{{{8}}}}\right)}}}+{\arctan{{\left({\frac{{{2}}}{{{25}}}}\right)}}}+{\arctan{{\left({\frac{{{1}}}{{{18}}}}\right)}}}+\ldots\right)}}}\)

Leia Sullivan

Leia Sullivan

Answered question

2022-04-06

I need to compute:
tan(arctan(12)+arctan(29)+arctan(18)+arctan(225)+arctan(118)+)

Answer & Explanation

chambasos6

chambasos6

Beginner2022-04-07Added 12 answers

Note that
n=1m(arctan(n+2)arctan(n))
=(arctan(3)arctan(1))+(arctan(4)arctan(2))+arctan(5)arctan(3)+
=arctanm+221+2(m+2)+arctanm+111+1(m+1)
=arctan12+5m+arctan11+2m
Hence we have
=arctan12+arctan1
Then, the answer will be
tan(arctan12+arctan1)=12+11121=3

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