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`underset("n-digits")((666 . . . .6)^(2))+underset("n-digits")((888 . . . .8))` is equal toA. `(4)/(9) (10^(n)-1)^(2)`B. `(4)/(9) (10^(n)-1)`C. `(4)/(9) (10^(2n)-1)`D. `(4)/(9) (10^(2n)-1)^(2)` |
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Answer» Correct Answer - C `(666............n" times")^(2)+ (888.............n" times ")` `implies (6+60+600+....n" times")^(2)+(8+80+800...n" times")` `implies{6[(10^(n)-1)/(10-1)]}^(2)+8[(10^(n)-1)/(10-1)]` ` implies(4)/(9)(10^(n)-1)^(2)+(8)/(9)(10^(n)-1)` `implies(4)/(9)(10^(n)-1){10^(n)-1+2}` ` implies (4)/(9) (10^(n)-1)(10^(n)+1)` `implies (4)/(9)(10^(2n)-1)` |
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