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Sum of n terms of the series 0.1 + 0.11 + 0.111 + … is
- a)1/9 {n – (1– (0.1)n)}
- b)1/9 {n – (1–(0.1)n)/9}
- c)n– 1 – (0.1)n/9
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
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Sum of n terms of the series 0.1 + 0.11 + 0.111 + … isa)1...
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Sum of n terms of the series 0.1 + 0.11 + 0.111 + … isa)1...
Given series: 0.1, 0.11, 0.111, ...
To find the sum of n terms of the series, we need to use the formula for the sum of a geometric series:
S = a(1 - r^n)/(1 - r)
where a is the first term, r is the common ratio, and n is the number of terms.
Let's find a and r for our series:
a = 0.1
r = 0.1 + 0.01 + 0.001 + ... = 1/10 + 1/100 + 1/1000 + ... = 1/9
Now, we can substitute these values into the formula and simplify:
S = 0.1(1 - (1/9)^n)/(1 - 1/9)
S = 0.1(9/8)(1 - (1/9)^n)
S = 9/80(1 - (1/10)^n)
So, the sum of n terms of the series is:
S = 1/9 n(1 - (0.1)^n)/9
Therefore, the correct answer is option B: 1/9 n(1 - (0.1)^n)/9.
To find the sum of n terms of the series, we need to use the formula for the sum of a geometric series:
S = a(1 - r^n)/(1 - r)
where a is the first term, r is the common ratio, and n is the number of terms.
Let's find a and r for our series:
a = 0.1
r = 0.1 + 0.01 + 0.001 + ... = 1/10 + 1/100 + 1/1000 + ... = 1/9
Now, we can substitute these values into the formula and simplify:
S = 0.1(1 - (1/9)^n)/(1 - 1/9)
S = 0.1(9/8)(1 - (1/9)^n)
S = 9/80(1 - (1/10)^n)
So, the sum of n terms of the series is:
S = 1/9 n(1 - (0.1)^n)/9
Therefore, the correct answer is option B: 1/9 n(1 - (0.1)^n)/9.
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Sum of n terms of the series 0.1 + 0.11 + 0.111 + … isa)1/9 {n – (1– (0.1)n)}b)1/9 {n – (1–(0.1)n)/9}c)n– 1 – (0.1)n/9d)none of theseCorrect answer is option 'B'. Can you explain this answer?
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