# 7 + 77 + 777 + ... + 777

Question:

$7+77+777+\ldots+777 \ldots \ldots \ldots \ldots 7=\frac{7}{81}\left(10^{n+1}-9 n-10\right)$

Solution:

Let P(n) be the given statement.

Now,

$P(n): 7+77+777+\ldots+777 \ldots n$ digits $\ldots 7=\frac{7}{81}\left(10^{n+1}-9 n-10\right)$

Step 1:

$P(1)=7=\frac{7}{81}\left(10^{2}-9-10\right)=\frac{7}{81} \times 81$

Thus, $P(1)$ is true.

Step 2 :

Let $P(m)$ be true.

Then,

$7+77+777+\ldots+777 \ldots m$ digits $\ldots 7=\frac{7}{81}\left(10^{m+1}-9 m-10\right)$

We need to show that $P(m+1)$ is true whenever $P(m)$ is true.

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

This is a geometric progression with $n=m+1$.

$\therefore S u m P(m+1):$

$=\frac{7}{9}[9+99+999+\ldots(m+1)$ term $]$

$=\frac{7}{9}[(10-1)+(100-1)+\ldots(m+1)$ term $]$

$=\frac{7}{9}[10+100+1000+\ldots(m+1)$ term $-(1+1+1 \ldots m+1$ times...+1 $]$

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

$=\frac{7}{81}\left[10^{m+2}-9 m-19\right]$

Thus, $P(m+1)$ is true.

By the principle of $m$ athematical $i$ nduction, $P(n)$ is true for all $n \in N$.