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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) 99A. 100B. 99C. 102D. 101 |
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Answer» 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`. |
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