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If `t_n`denotes the nth term of the series `2+3+6+11+18+...`then`t_(50)`=......`49^2-1`b. `49^2`c. `50^2+1`d. `49^2+2`A. `49^(2)-1`B. `49^(2)`C. `50^(2)+1`D. `49^(2)+2` |
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Answer» Correct Answer - D Let `S_(n)` be sum of the series 2+3+6+11+18+***+`t_(50).` `therefore s_(n) =2+3+6+11+18***+t_(50)` `and S_(n) =0+2+3+6+11+18***+t_(49) +y_(50)` On subtracting Eq (ii) from eq (i) we get `0=2+1+3+5+7+***` upto 49 terms `t_(50) =2+[1+3+55+7+***`upto 49 terms] `=2+(49)/(2)[2xx1++48xx2]` `=2+(49)/(2)xx[2+96]` `=2+[49+49xx48]` `2+49xx49=2+(49)^(2)` |
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