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If `S_(n)` denotes the sum to `n` terms of the series `(1 le n le 9)1 + 22 + 333 + "……" + ("nnn…..n")/(n"times")`, then for `n ge 2`A. `S_(n) - S_(n - 1) = (1)/(9)(10^(n) - n^(2) + n)`B. `S_(n) = (1)/(9)(10^(n) - n^(2) + 2n - 2)`C. `9(S_(n) - S_(n-1)) = n(10^(n) - 1)`D. `S_(3) = 356` |
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Answer» Correct Answer - C::D `{:(S_(n)=1+(2)/(9)(10^(2)-1)(3)/(9)(10^(3)-1)+"...."+(n)/(9)(10^(n)-1)),(S_(n-1)=1+(2)/(9)(10^(2)-1)(3)/(9)(10^(3)-1)+"..."+):}/(S_(n)-S_(n-1)=(n)/(9)(10^(n)-1))` `:.9(S_(n)-s_(n-1))=n(10^(n)-1)` `:. S_(3) = 356 , :. 3` & `4` |
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