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`S_(n)` be the sum of n terms of the series `(8)/(5)+(16)/(65)+(24)/(325)+"......"` The value of `S_(8)`, isA. `(288)/(145)`B. `(1088)/(545)`C. `(81)/(41)`D. `(107)/(245)` |
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Answer» Correct Answer - A Let `S_n=(8)/(5)+(16)/(65)+(24)/(325)+"......"` `T_(r)=(8r)/(4r^(4)+1)=(8r)/((2^(2)+2r+1)(2^(2)-2r+1))` `=2[((2^(2)+2r+1)-(2^(2)-2r+1))/((2^(2)+2r+1)(2^(2)-2r+1))]` `=2[(1)/((2^(2)-2r+1))-(1)/((2^(2)+2r+1))]` `S_(8)=sum_(r=1)^(8)T_(r)=2sum_(r=1)^(8)((1)/(2^(2)-2r+1)-(1)/(2^(2)+2r+1))` `=2(1-(1)/(2(8)^(2)+2(8)+1))=2(1-(1)/(145))=(288)/(145)` |
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