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Find the sum of n terms of series is (4-1/n)+(4-2/n)+(4-3/n)?
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Find the sum of n terms of series is (4-1/n)+(4-2/n)+(4-3/n)?
Understanding the Series
The series given is:
(4 - 1/n) + (4 - 2/n) + (4 - 3/n) + ... + (4 - n/n)
This can be expressed as a sum of two separate series:
- The constant part: 4 + 4 + 4 + ... (n terms)
- The variable part: (-1/n) + (-2/n) + (-3/n) + ... + (-n/n)
Sum of Constant Part
- The sum of the constant part is straightforward:
- Constant = 4
- Number of terms = n
- So, Sum = 4n
Sum of Variable Part
- The variable part can be simplified:
- This is an arithmetic series: -1/n, -2/n, -3/n, ..., -n/n
- The sum of the first n natural numbers is n(n + 1)/2.
- Therefore, Sum of the variable part = -(1/n)(1 + 2 + 3 + ... + n)
- This results in: -(1/n)(n(n + 1)/2) = -(n + 1)/2
Total Sum of n Terms
- Now, combine both sums:
- Total Sum = Sum of constant part + Sum of variable part
- Total Sum = 4n - (n + 1)/2
Final Expression
- The final expression for the sum of n terms is:
Total Sum = 4n - (n + 1)/2
This gives a clear understanding of how to derive the sum of the series for any n terms.
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Find the sum of n terms of series is (4-1/n)+(4-2/n)+(4-3/n)?
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