`underset("n-digits")((666 . . . .6)^(2))+underset("n-digits")((888 . . . .8))` is equal to

Options

A. `(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)`

Answer

C

Explanation

`(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)`