The sum of 4+44+444+.... upto n terms in

Given that, S = 4+44+444+...

$$\Rightarrow$$ $$S = \frac{4}{9}(9+99+999+...)$$

$$\Rightarrow$$ $$S = \frac{4}{9}(10-1+10^2-1+10^3-1+...+10^n-1)$$

$$\Rightarrow$$ $$S = \frac{4}{9}(10+10^2+10^3+...+10^n-n)$$

$$\Rightarrow$$ $$S = \frac{4}{9}(\frac{10(10^n - 1)}{10-1}-n)$$

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

Hence, option C is the correct answer. 

Alternate method: 

Solving for n = 2

Sum of series = 4+44 = 48

Substituting n = 2 in options

 (A) $$\frac{40}{81}(8^{2}-1)-\frac{5*2}{9}$$ = $$\frac{280-10}{9}$$ = 30

 (B) $$\frac{40}{81}(8^{2}-1)-\frac{4*2}{9}$$ = $$\frac{280-8}{9}$$ =  $$\frac{272}{9}$$

 (C) $$\frac{40}{81}(10^{2}-1)-\frac{4*2}{9}$$ = $$\frac{440-8}{9}$$ = 48

 (D) $$\frac{40}{81}(10^{2}-1)-\frac{5*2}{9}$$ = $$\frac{440-10}{9}$$ = $$\frac{430}{9}$$