Math, asked by manish6264, 1 year ago

find the sum upto n terms 0.3+0.33+0.333+...


manish6264: anyone solve this question?

Answers

Answered by Brendancrawford
0

Answer:


Step-by-step explanation:

3[0.1+0.11+0.111....0.111(ntimes)]  

39[0.9+0.99+0.999....0.999((ntimes)]

39[(1+1+1...)−(0.1+0.01....)]

39[(n)−1/10(1−1/10n)9/10]

39[(n)−1−1/10n9]

39[9n−1+1/10n9]

9n−1+1/10n27

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Ajinkya Kamath

Ajinkya Kamath, studied at St. Francis Institute of Technology

Answered Aug 31, 2017

Multiply and divide by 3,

The sequence becomes,

1/3(0.9+0.99+0.999+…n)

1/3(1–0.1+1–0.01+1–0.001+…n)

1/3(sum of all ones-(0.1+0.01+0.001…))

1/3(n-0.1(1–0.1^n)/(1–0.1))

1/3(n-1/9(1-0.1^n))

541 Views · View 1 Upvoter

Pradeep Kumar Singh

Pradeep Kumar Singh, B.Tech (M.E.) from Techno India College of Technology (2020)

Answered Aug 31, 2017

0.3+0.33+0.333+……n terms.

First we take 3 as common from the series,

So, 3×{0.1+0.11+0.111+…. n terms}

Now we multiply it by 9 and also divide it by 9 to neutralize the net value

3/9×{0.9+0.99+0.999+…..n terms}

3/9×{(1–0.1)+(1–0.01)+(1–0.001)+……n terms}

1 added upto n times to become ’n' , so we have

3/9×{n -(0.1+0.01+0.001+…… n terms) }

The term we get above (0.1+0.01+0.001+…..n terms) is in GP.

With a=0.1 & R= 0.1 so sum of the series is

Sn= a(1-r^n)/(1-r)

Sn=0.1(1–0.1^n)/(1–0.1)

Sn=(1–0.1^n)/9

So the some of the series is

3/9×{n-1/9(1–0.1^n) }

1/27×{9n-1+0.1^n}

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manish6264: the answer is n/3-1/27{(1-1/10^n)}
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