Question

Sum of the series , $7+77+777+.....$ to $n$ terms be $S=\left(\frac{7}{m}\right)(k^{n+1}-(k-1)n-k)$.Find $\frac{m-k-1}{k}$ ?

Answer 1

$7+77+777+...n=\frac{7}{9}[9+99+999+...n]$

$=\frac{7}{9}[(10-1)+({10}^2-1)+({10}^3-1)+...n]$

$=\frac{7}{9}[10+{10}^2+{10}^3+...n]-\frac{7}{9}[1+1+\mathrm{1...}n]$

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

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

$\therefore m=81,k=10$

$\therefore \frac{m-k-1}{k}=\frac{81-10-1}{10}=\frac{70}{10}=7$