Question

4 + 44 + 444 + .................upto $n$ terms = __

• A. $\frac{40}{81}\left({10}^n\right)+\frac{4n}{9}$
• B. $\frac{40}{81}({10}^n+1)+\frac{4n}{9}$
• C. $\frac{40}{81}\left({10}^n\right)-\frac{4n}{9}$
• D. $\frac{40}{81}({10}^n-1)-\frac{4n}{9}$

Answer 1

The correct option is D

$\frac{40}{81}({10}^n-1)-\frac{4n}{9}$

$4+44+444+.................\text{n terms}$

$=4(1+11+111+\cdots \cdots \text{n terms})$

$=\frac{4}{9}(9+99+999+\cdots \cdots \text{n terms})$

$=\frac{4}{9}\left[\right(10-1)+(100-1)+(1000-1)\cdots \cdots \text{n terms}]$

$=\frac{4}{9}\left[\right(10+{10}^2+{10}^3+\cdots \cdots \text{n terms})-(1+1+1\cdots \cdots \text{n terms}\left)\right]$

$=\frac{4}{9}[\frac{10({10}^n-1)}{10-1}-n]$

$=\frac{40}{81}({10}^n-1)-\frac{4n}{9}$