Question

Find the sum of the series 0.4+0.44+0.444+----- to n terms

Answer 1

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Answer 2

Answer:

The sum of the series is

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

Step-by-step explanation:

Given : Series 0.4+0.44+0.444+----- to n terms

To find : The sum of the series?

Solution :

Let the series be 'x'

$x=0.4+0.44+0.444+...+ \text{n terms}$

$x=4(0.1+0.11+0.111+...+ \text{n terms})$

Multiply and divide by 9,

$x=\frac{4}{9}(0.9+0.99+0.999+..+ \text{n terms})$

$x=\frac{4}{9}[(1-0.1)+(1-0.01)+(1-0.001)...+ \text{n terms}]$

$x=\frac{4}{9}[(1-0.1)+(1-(0.1)^2)+(1-(0.1)^3)...+ \text{n terms}]$

$x=\frac{4}{9}[(1+1+1+...\text{n terms})-(0.1+(0.1)^2+(0.1)^3...+ \text{n terms})]$

$x=\frac{4}{9}[(n)-(0.1(\frac{1-(0.1)^n}{1-0.1}))]$

$x=\frac{4}{9}[n-\frac{1-(0.1)^n}{9}]$

Therefore, The sum of the series is $0.4+0.44+0.444+...+ \text{n terms}=\frac{4}{9}[n-\frac{1-(0.1)^n}{9}]$