Question

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
A. 100
B. 99
C. 102
D. 101

Answer 1

Correct Answer - D
`((8)/(5))^(2)+((12)/(5))^(2)+((16)/(5))^(2)+((20)/(5))^(2)+"......."+((44)/(5))^(2)`
`=(16)/(25)(2^(2)+3^(2)+4^(2)+5^(2)+"........"+11^(2))`
`=(16)/(25)((11*(11+1)*(22+1))/(6)-1)`
`=(16)/(25)xx505=(16)/(25)xx101=(16)/(5)m(" given ")`
`:.m =101`.