Sum of series ∑ r = 1 n sin − 1 [ r ( r + 1 ) ( r 2 + 2 r + r 2 − 1 ) 2 r + 1 ] is

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Sum of series ∑ r = 1 n ​ sin − 1 [ r ( r + 1 ) ( r 2 + 2 r ​ + r 2 − 1 ​ ) 2 r + 1 ​ ] is

TestHub · Accepted Answer

Since 2 r + 1 = ( r 2 + 2 r ) − ( r 2 − 1 ) : ​ T r ​ = sin − 1 ( r ( r + 1 ) r 2 + 2 r ​ − r 2 − 1 ​ ​ ) = sin − 1 ( r 1 ​ 1 − ( r + 1 ) 2 1 ​ ​ − r + 1 1 ​ 1 − r 2 1 ​ ​ ) = sin − 1 r 1 ​ − sin − 1 r + 1 1 ​ ​ S n ​ = 2 π ​ − sin − 1 n + 1 1 ​ = cos − 1 n + 1 1 ​ ​

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