Question

find the sum to n yerms of the series o.4+o.44+o.444...to n terms​

Answer 1

Answer:

0.4 + 0.44 + 0.444 + 0.44...(upto n terms) = $\frac{4}{9}(n - (\frac{1 - (0.1)^n}{9} )$

Step-by-step explanation:

Given :

To find the sum to n yerms of the series 0.4 + 0.44 + 0.444...to n terms​,.

Solution :

⇒ 0.4 + 0.44 + 0.444...to n terms

⇒ 4 (0.1 + 0.11 + 0.111...)

multiplying & dividing by 9,

We get,

⇒ $\frac{4}{9}(9) (0.1 + 0.11 + 0.111...)$

⇒ $\frac{4}{9}(0.9 + 0.99 + 0.999...)$

⇒ $\frac{4}{9}((1 - 0.1) + (1 - 0.01) + ...)$ n terms

⇒  $\frac{4}{9}((1 - 0.1) + (1 - (0.1)^2) +(1-(0.1)^3) + ...)$ n terms

⇒ $\frac{4}{9}(n - ( (0.1) + (0.1)^2 + (0.1)^3 + ...)$

By using the formula,

$S_n = a\frac{1-r^n}{1-r}$

⇒ $\frac{4}{9}(n - ( (0.1) + (0.1)^2 + (0.1)^3 + ...)$

⇒ $\frac{4}{9}(n - ((0.1)\frac{1 - (0.1)^n}{1 - 0.1} )$

⇒ $\frac{4}{9}(n - ((0.1)\frac{1 - (0.1)^n}{0.9} )$

⇒ $\frac{4}{9}(n - (\frac{1 - (0.1)^n}{9} )$