If `Sigma_(r=1)^(2n) sin^(-1) x^(r )=n pi, then Sigma__(r=1)^(2n) x_(r )` is equal toA. nB. 2nC. `(n(n+1))/(2)`D. none of these

If `Sigma_(r=1)^(2n) sin^(-1) x^(r )=n pi, then Sigma__(r=1)^(2n) x_(r

1.	If `Sigma_(r=1)^(2n) sin^(-1) x^(r )=n pi, then Sigma__(r=1)^(2n) x_(r )` is equal toA. nB. 2nC. `(n(n+1))/(2)`D. none of these
Answer» We have `(pi)/(2)lt sin^(-1)x_(e)le(pi)/(2)for i=1,2…,2n` `therefore underset(r=1)overset(2n)Sigma sin^(-1)x_(r )=npi` `rarr sin^(-1) pi_(r ),r=1,2..,2n` `rarr underset(r=1)overset(2n)Sigma x_(r )=underset(r=1)overset(2n)Sigma 1=2n`

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