(1)/(4)" and "(1)/(2)

(1)/(4)" " (1)/(2) " "(3)/(4)" "1" "1(1)/(4)" "1(1)/(2)" "1(3)/(4) ?

Simplify : 8 (1)/(2)-[3 (1)/(4)-:{1(1)/(4) - (1)/(2) ( 1 (1)/(2)-(1)/(3)-(1)/(6))}]

The sum of the first 35 terms of the series (1)/(2)+(1)/(3)-(1)/(4)-(1)/(2)-(1)/(3)+(1)/(4)+(1)/(2)+(1)/(3)-(1)/(4) is -(1)/(2)(b)-(1)/(4) (c) (1)/(4) (e) None of these

Prove that "tan"^(-1)(1)/(4) +"tan"^(-1)(2)/(9) =(1)/(2)"tan"^(-1)(4)/(3) .

The sum of the series 1 + (1)/(1!) ((1)/(4)) + (1.3)/(2!) ((1)/(4))^(2) + (1.3.5)/(3!) ((1)/(4))^(3)+ ... to infty , is

The sum of the series 1 + (1)/(1!) ((1)/(4)) + (1.3)/(2!) ((1)/(4))^(2) + (1.3.5)/(3!) ((1)/(4))^(3)+ ... to infty , is

3log 2 +(1)/(4) -(1)/(2)((1)/(4))^(2)+(1)/(3)((1)/(4))^(3)-….. =