Question

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 `16/5m` ,then m is equal to
A. 102
B. 101
C. 100
D. 99

Answer 1

Correct Answer - B
Let `S_(10)` be the sum of first ten terms of the series. Then, we have
`S_(10) = (1 (3)/(5))^(2) + (2(2)/(5))^(2) + (3(1)/(5))^(2) + 4^(2) + (4(4)/(5))^(2)` + ... to 10 terms
`= ((8)/(5))^(2) + ((12)/(5))^(2) + ((16)/(5))^(2) + 4^(2) + ((24)/(5))^(2) +`....to 10 terms
`= (1)/(5^(2)) (8^(2) + 12^(2) + 16^(2) + 20^(2) + 24^(2) + .. " to 10 terms")`
`= (4)/(5^(2)) (2^(2) + 3^(2) + 4^(2) + 5^(2) + ..." to 10 terms")`
`= (4^(2))/(5^(2)) (2^(2) + 3^(2) + 4^(2) + 5^(2) + ...+ 11^(2))`
`= (16)/(25) ((1^(2) + 2^(2) + ... + 11^(2)) -1^(2))`
`= (16)/(25) ((11.(11 + 1) (2 .11 + 1))/(6) -1)`
`= (16)/(25) (506 -1) = (16)/(25) xx 505 rArr (16)//(5)m = (16)/(25) xx 505 = 101`