show that `(7)/(2^((1)/(2)) + 2^((1)/(4)) + 1) = 1 - 3.2^((1)/(4)) + 2.2^((1)/(2)) + 2^((3)/(4))`

(7)/(2^((1)/(2))+2^((1)/(4))+1)=A+B*2^((1)/(4))+C.2^((1)/(2))+D.2^((3)/(4))

Solve : (i) 2^((2)/(3))*2^((1)/(4)) (ii) ((1)/(3^(3)))^(7) (iii) (11^((1)/(2)))/(11^((1)/(4))) (iv) 7^((1)/(2))*8^((1)/(2))

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

The sum of the series 1 + (1)/(3^(2)) + (1 *4)/(1*2) (1)/(3^(4))+( 1 * 4 * 7)/(1 *2*3)(1)/(3^(6)) + ..., is (a) ((3)/(2))^((1)/(3)) (b) ((5)/(4))^((1)/(3)) (c) ((3)/(2))^((1)/(6)) (d) None of these

[{((1)/(7^(2)))^(-2)}^((1)/(3))]^((1)/(4))=7^(m), is -(1)/(3)(b)(1)/(4)(c)-3(d)2

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

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

7(1)/(2)-((8)/(11)xx7(1)/(3))-:(3(1)/(2)-2(1)/(4))