Question

Prove that $7+77+777+......+777\underset{n-digits}{.............}7=\frac{7}{81}({10}^{n+1}-9n-10)$

Answer 1

Let P(n) : $7+77+777+.....+777+\underset{n-digits}{.............}+7=\frac{7}{81}({10}^{n+1}-9n-10)$

For n = 1

$7=\frac{7}{81}({10}^2-9-10)$

$7=\frac{7}{81}(100-19)$

$7=7$

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k, so

$7+77+777+......+777\underset{n-digits}{.............}7=\frac{7}{81}$

${10}^{k+1}-9k-10$ ......(1)

We have to show that,

$7+77+777+...+7777\underset{n-digits}{.............}7+777\underset{(k+1)digits}{.............}7=\frac{7}{81}[{10}^{k+2}-9(k-1)-10]$

Now,

$\{7+77+777+.....+777\underset{k-digits}{.............}7\}+777\underset{(k+1)-digits}{.............}7$

$=\frac{7}{81}[{10}^{k+1}-9k-10]+\frac{7}{9}({10}^{k+1}-1)$

[Using equation (1)]

$=7[\frac{{10}^{k+1}-9k-10}{81}+\frac{1}{9}({10}^{k+1}-1\left)\right]$

$=\frac{7}{81}[{10}^{k+1}-9k-10+{9.10}^{k+1}-9]$

$=\frac{7}{81}\left[{10}^{k+1}\right(9+1)-9(k+1)-10]$

$\Rightarrow$ P(n) is true for n = k + 1

$\Rightarrow$ P(n) is true for all $nϵN$ by PMI.