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The value of `sum_(r=1)^oo tan^-1 (4/(4r^2 +3))=`A. 1B. 3C. 2D. 4 |
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Answer» Correct Answer - C let `T_(r)=tan^(-1)((4)/(4r^(2)+3))=tan^(-1)((1)/((3)/(4)+r^(2)))` `=tan^(-1)((1)/(1+r^(2)-(1)/(4)))` `=tan^(-1)(((r+(1)/(2))-(r-(1)/(2)))/(1+(r+(1)/(2))(r-(1)/(2))))` `=tan^(-1)(r+(1)/(2))-tan^(-1)(r-(1)/(2))` `thereforesum_(r=1)^(infty)T_(r)=tan^(_1)((3)/(2))-tan^(-1)((1)/(2))` `+tan^(-1)((5)/(2))-tan^(-1)((3)/(2))` `+tan^(-1)((7)/(2))-tan^(-1)((5)/(2))` `thereforesum_(r=1)^(infty)T_(r)=(pi)/(2)-tan^(-1)((1)/(2))=cot^(-1)((1)/(2))=tan^(-1)(2)` `thereforetan(sum_(r=1)^(infty)T_(r))=2` |
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