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

Explanation

`S_(1)=underset(n "digits")(6666…)`
`=6+6xx10^(1)+6xx10^(2)+….+6xx10^(n-1)`
`=6xx((10^(n)-1))/(10-1)=2/3(10^(n)-1)`
Similarly,
`S_(2)=8/9(10^(n)-1)`
`rArrS_(1)^(2)+S_(2)=4/9(10^(n)-1)^(2)+8/9(10^(n)-1)`
`=4/9(10^(n)-1)[10^(n)-1+2]`
`=4/9[10^(2n)-1]`
Also, `underset(2n "digits")(4444...4)=4+$xx10+4xx10^(2)+...+4xx10^(2n-1)`
`=4((10^(2n)-1))/(10-1)`
Hence, proved.