If the surm of the first ten terms of the series,`(1 3/5)^2+(2 2/5)^2+ – InterviewSolution

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1.	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 toA. 101B. 100C. 99D. 102
Answer» Correct Answer - A `S_n=(8/5)^5+(12/5)^2+(16/5)^2+(20/5)^2+(24/5)^4+...` `=8^2/5^2+12^2/5^2+16^2/5^2+20^2/5^2+24^2/5+.....` `therefore T_n=(4n+4)^2/5^2` `therefore S_n =1/5^2 underset(n=1)overset(10)Sigma16 (n+1)^2` `=16/25[(underset(n=1)overset(11)Sigman^2)-1]` `=16/25[(11xx12xx23)/(6)-1]` `=16/25xx 505 =16/5m` `rArr m= 101`

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