prove that `7 + 77 + 777 +...... + 777........._(n-digits) 7 = 7/81 (10^(n+1) - 9n - 10)` for all `n in N`

Answer

C

Explanation

`because ubrace(777......7)_("n digits")=7ubrace((111......1))_("n digits")`
`=7(1+10+10^2+....+10^(n-1))`
`=7+7xx10+7xx10^2+......+7xx10^(n-1)`
`ne 7+7xx10+7xx10^(2)+....+7xx10^n)`
`therefore` Statement -2 is false .
Now ,let `P(n):7+77+777+.....+ubrace(777......7)_("n digits")=(7)/(81)(10^(n+1)-9n-10)`
Step I For n=1,
LHS =7 and RHS `=(7)/(81)(10^2-9-10)=7`
`therefore LHS=RHS`
which is true for `n=1`.
Step II Assume P(n) is true for n=k, then
`P(k):7+77+777+.....+ubrace(777......7)_("k digits")=(7)/(81)(10^(k+1)-9k-10)`
Step III For `n=k+1`.
`P(k+1):7+77+777+....+ubrace(777......7)_("n digits")+ubrace(777.7)_((k+1)digits)=`
`(7)/(81)[10^(k+2)-9(k+1)-10]`
LHS =7+77+777+.....+ubrace(777......7)_("k digits")+ubrace(777......7)_((k+1)digits)`
`=(7)/(81)(10^(k+1)-9k-10)+7(1+10+10^2+......+10^k)`
`=(7)/(81)(10^(k+1)-9k-10)+(7(1k^(k+1)-1))/(10-1)`
`=(7)/(81)(10^(k+1)-9k-10+9.10^(k+1)-9)`
`=(7)/(81)[10^(k+1)(1+9)-9(k+1-10]`
`=(7)/(81)[10^(k+2-9(k+1-10]`
`=RHS`
Therefore , `P(k+1)` is true . Hence , by mathematical induction `P(n)` is true for all natural numbers .
Hence , Statement-1is true and Statement -2 is false .