(3)/(5)+(4)/(5^(2))+(3)/(5^(3))+(4)/(5^(4))+..." to "2n" terms; "

The sum of 1+(2)/(5)+(3)/(5^(2))+(4)/(5^(3))+.... upto n terms

Find the sum of the series upto infinite terms (3)/(4)-(5)/(4^(2))+(3)/(4^(3))-(5)/(4^(4))+...

-((4)/(5)+(4^(2))/(5^(2).2)+(4^(3))/(5^(3).3)+(4^(4))/(5^(4).4)+……oo)=

-((4)/(5)+(4^(2))/(5^(2).2)+(4^(3))/(5^(3).3)+(4^(4))/(5^(4).4)+……oo)=

The value of (9)/(15) " of " ((2)/(3) // (2)/(3) " of " (3)/(2)) ÷ ((3)/(4) xx (3)/(4) ÷ (3)/(4) " of " (4)/(3)) " of " ((5)/(4) ÷ (5)/(2) xx (2)/(5) " of " (4)/(5)) is :

The value of (9)/(15) " of " ((2)/(3) // (2)/(3) " of " (3)/(2)) ÷ ((3)/(4) xx (3)/(4) ÷ (3)/(4) " of " (4)/(3)) " of " ((5)/(4) ÷ (5)/(2) xx (2)/(5) " of " (4)/(5)) is :

If the surm of the first ten terms of the series,(1(3)/(5))^(2)+(2(2)/(5))^(2)+(3(1)/(5))^(2)+4^(2)+(4(4)/(5))^(2)+ is (16)/(5)m, then m is equal to

(3)/(4)-(5)/(4^(2))+(3)/(4^(3))-(5)/(4^(4))+(3)/(4^(5))-(5)/(4^(6))+.....oo= ?

If A=[(2)/(3)1(5)/(3)(1)/(3)(2)/(3)(4)/(3)(7)/(3)2(2)/(3)] and B=[(2)/(3)(2)/(5)1(1)/(5)(2)/(5)(4)/(5)(4)/(5)(7)/(3)(6)/(5)(2)/(5)], then compute 3A_(-)=5B