Question

Sum of the series $4+44+444+...$ to $n$ terms is

• A. $\frac{4n}{9}-\frac{4}{9^2}(1-\frac{1}{{10}^n})$
• B. $\frac{4n}{81}-\frac{4}{729}(1-\frac{1}{{10}^n})$
• C. $\frac{4n}{81}+\frac{4}{9^2}(1-\frac{1}{{10}^n})$
• D. None of these

Answer 1

The correct option is D None of these

$S_n=4+44+444+................n-terms\Rightarrow \frac{9}{4}{S}_n=9+99+999+.........\Rightarrow \frac{9}{4}{S}_n=(10-1)+(100-1)+(1000-1)+.....\Rightarrow \frac{9}{4}{S}_n=10+100+1000+.......n-terms-n=\frac{10({10}^n-1)}{9}-n\Rightarrow S_n=\frac{40({10}^n-1)}{81}-\frac{4n}{9}$
The correct answer is D