lf n∈N,k=1∑nsin−1(xk)=2nπ then k=1∑nxk=

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lf {n} {N},_{k=1}^{n}^{-1}(x_{k})={n}{2} then _{k=1}^{n}x_{k

Question

lf $n \in N, k = 1 \sum n sin^{- 1} (x_{k}) = \frac{nπ}{2}$ then $k = 1 \sum n x_{k} =$

n

k

\frac{k ( k + 1 )}{2}

\frac{n ( n + 1 )}{2}

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Michael Discussed

n \in N, k = 1 \sum n sin^{- 1} (x_{k}) = \frac{nπ}{2}

then

k = 1 \sum n x_{k} =

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Question Text

n \in N, k = 1 \sum n sin^{- 1} (x_{k}) = \frac{nπ}{2}

then

k = 1 \sum n x_{k} =