The Problem is 2i^2 + 5. Starting with i = 1 and ending with i = 50. I need to find the partial sum but I don't want to just add everything with the series from 1 to 50
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This can be broken down to: 2(Sum)i^2 + (Sum)5 (both 1 to 50)
The sum of the power series is given by (n(n + 1)(2n + 1))/6; for n = 50,
sum = 50(51)(101)/6 = 42925; multiply by two, add 250 = 86100.
The sum of the power series is given by (n(n + 1)(2n + 1))/6; for n = 50,
sum = 50(51)(101)/6 = 42925; multiply by two, add 250 = 86100.
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2i² + 5 =
2 * [(i+1)³ - i³ - 3i - 1]/3 + 5 =
2/3 * [(i+1)³ - i³] - 2*i + 4+1/3 =
2/3 * [(i+1)³ - i³] - [(i+1)² - i² - 1] + 4+1/3 =
2/3 * [(i+1)³ - i³] - [(i+1)² - i²] + [5+1/3]
Now it should be evident how to evaluate the sum:
2/3 * (51³ - 1) - (51² - 1) + (5+1/3)*50 = 86100
You can read about the calculus of finite differences in the book 'Concrete Mathematics', for example.
2 * [(i+1)³ - i³ - 3i - 1]/3 + 5 =
2/3 * [(i+1)³ - i³] - 2*i + 4+1/3 =
2/3 * [(i+1)³ - i³] - [(i+1)² - i² - 1] + 4+1/3 =
2/3 * [(i+1)³ - i³] - [(i+1)² - i²] + [5+1/3]
Now it should be evident how to evaluate the sum:
2/3 * (51³ - 1) - (51² - 1) + (5+1/3)*50 = 86100
You can read about the calculus of finite differences in the book 'Concrete Mathematics', for example.