Question

Find the sum to $n$ terms of the series
$0.7+7.7+0.77+77.7+0.777+777.7+0.7777+....$ where $n$ is even.

• A. $\frac{7}{9}\left\{\frac{n}{4}-\frac{(1-{\left(0.1\right)}^{n/4})}{9}\right\}+\frac{7}{9}\left\{\frac{10\left(\right(10{)}^{n/4}-1)}{9}-0.1\frac{n}{4}\right\}$
• B. $\frac{7}{9}\left\{\frac{n}{2}-\frac{(1-{\left(0.1\right)}^{n/2})}{9}\right\}+\frac{7}{9}\left\{\frac{10\left(\right(10{)}^{n/2}-1)}{9}-\frac{n}{2}\right\}$
• C. $\frac{7}{9}\left\{\frac{n}{2}+\frac{(1-{\left(0.1\right)}^{n/2})}{9}\right\}+\frac{7}{9}\left\{\frac{10\left(\right(10{)}^{n/2}-1)}{9}+\frac{n}{2}\right\}$
• D. none of these

Answer 1

The correct option is B $\frac{7}{9}\left\{\frac{n}{2}-\frac{(1-{\left(0.1\right)}^{n/2})}{9}\right\}+\frac{7}{9}\left\{\frac{10\left(\right(10{)}^{n/2}-1)}{9}-\frac{n}{2}\right\}$

Let $S=0.7+7.7+0.77+77.7+0.777+777.7+0.7777+...$.
We have to find the sum of $n$ terms of the series, where $n$ is even.
Assume, $n=2m$.
$\therefore S=(0.7+0.77+0.777+...mterms)+(7.7+77.7+777.7+...mterms)$
$\therefore S=\frac{7}{9}(0.9+0.99+0.999+...mterms)+\frac{7}{90}\left[\right({10}^2-1)+({10}^3-1)+...+({10}^{m+1}-1\left)\right]$
$\therefore S=\frac{7}{9}\{\frac{n}{2}-\frac{1}{9}(1-\frac{1}{{10}^{\frac{n}{2}}})\}+\frac{7}{90}\left\{\frac{100}{9}\right({10}^{\frac{n}{2}}-1)-\frac{n}{2}\}$
$\therefore S=\frac{7}{9}\{\frac{n}{2}-\frac{(1-{\left(0.1\right)}^{\frac{n}{2}})}{9}\}+\frac{7}{9}\{\frac{10\left(\right(10{)}^{\frac{n}{2}}-1)}{9}-\frac{n}{2}\}$