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If `(10)^(9)+2(11)^(1)(10)^(8)+3(11)^(2)(10)^(7)+"........"+(10)(11)^(9)=k(10)^(9)`, then k is equal toA. 100B. 110C. `(121)/(10)`D. `(441)/(100)` |
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Answer» Correct Answer - A The given series can be written as `k = 1+2(11/10)+3(11/10)^2""+...+9(11/10)^8""+10(11/10)^9" ...(i)"` On multiplying both sides by `(11/10)`, then `(11k)/10=(11/10)+2(11/10)^2 +3(11/10)^3""+...+9(11/10)^9""+10(11/10)^10" ...(ii)"` Now, on subtracting Eq. (ii) frm Eq. (i), then `-k/10=(11/10)+(11/10)^2""+...+(11/10)^2-10(11/10)^10` `=(1*{(11/10)^10-1})/((11/10-1))-10(11/10)^10` `rArr" "k = -100*{(11/10)^10-1}+100(11/10)^10 = 100` |
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