If the sum 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: (1) 102 (2) 101 (3) 100 (4) 99

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

The sum of the first n terms of the series 1^(2)+2.2^(2)+3^(2)+2.4^(2)+5^(2)+2.6^(2)+... is

Find the sum to n terms of the series 1.2.3+2.3.4+3.4.5+…

The sum first 19 terms of the series 1^2+2(2^2)+3^2+2(4^2)+5^2+... is :

Find the sum of n terms of the series 1+(4)/(5)+(7)/(5^(2))+(10)/(5^(3))+

If the sum of first 15 terms of series ((4)/(3))^(3)+(1(3)/(5))^(3)+(2(2)/(5))^(3)+(3(1)/(5))^(3)+4^(3) is equal to (512)/(2)k then k is equal to:

Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series 1^(2)+2.2^(2)+3^(2)+2.4^(2)+5^(2)+2.6^(2)+... If B-2A=100 lambda then lambda is equal to (1)232(2)248 (3) 464 (4) 496