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Sum of n terms of the series 4 + 44 + 444 + … is
- a)4/9 {10/9 (10n –1) –n}
- b)10/9 (10n –1) –n
- c)4/9 (10n –1) –n
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
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Sum of n terms of the series 4 + 44 + 444 + … isa)4/9 {10/9 (10n...
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Sum of n terms of the series 4 + 44 + 444 + … isa)4/9 {10/9 (10n...
Given series: 4, 44, 444, ...
To find: Sum of n terms of the series
Approach:
- We can observe that each term of the series is formed by concatenating the digit 4, n times.
- For example, the first term is 4, the second term is 4 followed by another 4 (i.e., 44), the third term is 4 followed by two 4's (i.e., 444), and so on.
- Therefore, the ith term of the series can be written as 4 + 40 + 400 + ... (i-1 times) + 4...4 (n-i times).
- Simplifying this expression, we get: ith term = 4(10^(i-1)) + 4...4 (n-i times).
- Now, we can use the formula for sum of n terms of an arithmetic progression to find the sum of the given series.
Formula: Sum of n terms of an arithmetic progression = (n/2) x (first term + last term)
- In our case, the first term of the series is 4 and the common difference is 40.
- Therefore, the last term of the series can be obtained by substituting i=n in the expression for ith term:
- nth term = 4(10^(n-1)) + 4...4 (n-n times) = 4(10^(n-1))
- Using the formula for sum of n terms, we get:
Sum of n terms = (n/2) x (first term + last term)
= (n/2) x (4 + 4(10^(n-1)))
= 2n(1 + 10^(n-1))
- Simplifying further, we get:
Sum of n terms = (2/9) x (10n - 1)
- Therefore, the correct option is A: 4/9 x (10n - 1)/(10 - 1) = 4/9 x (10n - 1)
To find: Sum of n terms of the series
Approach:
- We can observe that each term of the series is formed by concatenating the digit 4, n times.
- For example, the first term is 4, the second term is 4 followed by another 4 (i.e., 44), the third term is 4 followed by two 4's (i.e., 444), and so on.
- Therefore, the ith term of the series can be written as 4 + 40 + 400 + ... (i-1 times) + 4...4 (n-i times).
- Simplifying this expression, we get: ith term = 4(10^(i-1)) + 4...4 (n-i times).
- Now, we can use the formula for sum of n terms of an arithmetic progression to find the sum of the given series.
Formula: Sum of n terms of an arithmetic progression = (n/2) x (first term + last term)
- In our case, the first term of the series is 4 and the common difference is 40.
- Therefore, the last term of the series can be obtained by substituting i=n in the expression for ith term:
- nth term = 4(10^(n-1)) + 4...4 (n-n times) = 4(10^(n-1))
- Using the formula for sum of n terms, we get:
Sum of n terms = (n/2) x (first term + last term)
= (n/2) x (4 + 4(10^(n-1)))
= 2n(1 + 10^(n-1))
- Simplifying further, we get:
Sum of n terms = (2/9) x (10n - 1)
- Therefore, the correct option is A: 4/9 x (10n - 1)/(10 - 1) = 4/9 x (10n - 1)
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Sum of n terms of the series 4 + 44 + 444 + … isa)4/9 {10/9 (10n...
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Sum of n terms of the series 4 + 44 + 444 + … isa)4/9 {10/9 (10n –1) –n}b)10/9 (10n –1) –nc)4/9 (10n –1) –nd)none of theseCorrect answer is option 'A'. Can you explain this answer?
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