Question

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

Answer 1

Let P(n) be the given statement.
Now,
$$\begin{gathered} P\left(n\right):7+77+777+...+777..{.}_{n\mathrm{digits}}...7=\frac{7}{81}({10}^{n+1}-9n-10) \\ \mathrm{Step}\left(1\right): \\ P\left(1\right)=7=\frac{7}{81}({10}^2-9-10)=\frac{7}{81}\times 81 \\ \mathrm{Thus},P\left(1\right)\mathrm{is}\mathrm{true}. \\ \mathrm{Step}2: \\ \mathrm{Let}P\left(m\right)be\mathrm{true}. \\ \mathrm{Then}, \\ 7+77+777+...+777..{.}_{mdigits}...7=\frac{7}{81}({10}^{m+1}-9m-10) \\ \mathrm{We}\mathrm{need}\mathrm{to}\mathrm{show}\mathrm{that}P(m+1)\mathrm{is}\mathrm{true}\mathrm{whenever}P\left(m\right)\mathrm{is}\mathrm{true}. \end{gathered}$$

Now, P(m + 1) = 7 + 77 + 777 +....+ 777...(m + 1) digits...7

$$\begin{gathered} \mathrm{This}\mathrm{is}\mathrm{a}\mathrm{geometric}\mathrm{progression}\mathrm{with}n=m+1. \\ \therefore S\mathrm{um}P(m+1): \\ =\frac{7}{9}\left[9+99+999+...\left(m+1\right)\mathrm{term}\right] \\ =\frac{7}{9}\left[\left(10-1\right)+\left(100-1\right)+...(m+1)\mathrm{term}\right] \\ =\frac{7}{9}\left[10+100+1000+...(m+1)\mathrm{term}-(1+1+1...m+1times...+1\right] \\ =\frac{7}{9}\left[\frac{10\left({10}^{m+1}-1\right)}{9}-m+1\right] \\ =\frac{7}{81}\left[{10}^{m+2}-9m-19\right] \\ \mathrm{Thus},P(m+1)\mathrm{is}\mathrm{true}. \\ Bythep\mathrm{rinciple}\mathrm{of}m\mathrm{athematical}i\mathrm{nduction},P\left(n\right)\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{all}n\in N. \end{gathered}$$