Sum year we're having
If k = 1 ∑ 2 0 1 5 cot − 1 ( 2 k 2 ) = cot − 1 ( b a ) , where a and b are positive coprime integers, then find a + b .
The answer is 4031.
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2 solutions
Perfect Solution. Did the exact same way.
And just to add another way telescoping that -
cot − 1 ( 2 k 2 ) = tan − 1 ( 4 k 2 2 ) = tan − 1 ( 1 + ( 4 k 2 − 1 ) 2 ) = tan − 1 ( 1 + ( 2 k + 1 ) ( 2 k − 1 ) ( 2 k + 1 ) − ( 2 k − 1 ) ) = tan − 1 ( 2 k + 1 ) − tan − 1 ( 2 k − 1 )
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Nice. So we end up with
tan − 1 ( 4 0 3 1 ) − tan − 1 ( 1 ) =
tan − 1 ( 1 + 4 0 3 1 4 0 3 1 − 1 ) = tan − 1 ( 2 0 1 6 2 0 1 5 ) = cot − 1 ( 2 0 1 5 2 0 1 6 ) .
I actually prefer your way of telescoping. :)
cot^-1 (2k^2) = tan^-1 (1/2k^2)= tan^-1 (2k+1) -tan^-1 (2k-1) therefore the summation becomes , tan^-1 (4031) -tan^-1 (1) =.......
Note that cot − 1 ( 2 k 2 ) = cot − 1 ( 1 + k 1 ) − cot − 1 ( 1 + k − 1 1 ) .
This follows from the identity cot − 1 ( a ) − cot − 1 ( b ) = cot − 1 ( b − a a b + 1 ) .
So we end up with a telescoping series with only
− cot − 1 ( ∞ ) + cot − 1 ( 1 + 2 0 1 5 1 ) = 0 + cot − 1 ( 2 0 1 5 2 0 1 6 )
remaining. Thus a + b = 2 0 1 6 + 2 0 1 5 = 4 0 3 1 .