What is the sum of 7+77+777+7777+... to n terms ?

Nov 7, 2016

${\sum}_{k = 1}^{n} {a}_{k} = \frac{70}{81} \left({10}^{n} - 1\right) - \frac{7}{9} n$

Explanation:

Note that $\frac{7}{9} = 0. \overline{7}$

Hence we can write a formula for the $k$th term:

${a}_{k} = \frac{7}{9} \left({10}^{k} - 1\right) \text{ }$ for $k = 1 , 2 , 3. . .$

Let:

${b}_{k} = \frac{7}{9} \left({10}^{k}\right) = \frac{70}{9} \cdot {10}^{k - 1}$

This is in the form ${b}_{k} = b \cdot {r}^{k - 1}$, the general term of a geometric series with initial term $b = \frac{70}{9}$ and common ratio $r = 10$

The sum to $n$ terms is given by the formula:

${S}_{n} = \frac{b \left({r}^{n} - 1\right)}{r - 1} = \frac{70}{9} \cdot \frac{{10}^{n} - 1}{10 - 1} = \frac{70}{81} \cdot \left({10}^{n} - 1\right)$

Hence:

${\sum}_{k = 1}^{n} {a}_{k} = \frac{70}{81} \left({10}^{n} - 1\right) - \frac{7}{9} n$