9+11+13+15+cdots+99=sum_{k=0}^{45} square

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Accepted Answer

Step:1 ：Define the first term (a) as 9, the last term (an) as 99, and the common difference (d) as 2.Step:2 ：Calculate the number of terms (n) using the formula (n = frac{{an - a}}{d} + 1). Substituting the values, we get (n = frac{{99 - 9}}{2} + 1 = 46.0).Step:3 ：Calculate the sum (S) of the series using the formula (S = frac{n}{2} cdot (a + an)). Substituting the values, we get (S = frac{46.0}{2} cdot (9 + 99) = 2484.0).Step:4 ：Since the sum of an arithmetic series should be an integer, we can conclude that the sum is 2484.Step:5 ：Final Answer: (boxed{2484})

Answer

Step: 1 ：Define the first term (a) as 9, the last term (an) as 99, and the common difference (d) as 2.Step: 2 ：Calculate the number of terms (n) using the formula (n = frac{{an - a}}{d} + 1). Substituting the values, we get (n = frac{{99 - 9}}{2} + 1 = 46.0).Step: 3 ：Calculate the sum (S) of the series using the formula (S = frac{n}{2} cdot (a + an)). Substituting the values, we get (S = frac{46.0}{2} cdot (9 + 99) = 2484.0).Step: 4 ：Since the sum of an arithmetic series should be an integer, we can conclude that the sum is 2484.Step: 5 ：Final Answer: (boxed{2484})

$9 + 11 + 13 + 15 + \dots + 99 = \sum_{k = 0}^{45} ◻$

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Popular Problems

9+11+13+15+⋯+99=∑k=045◻

Answer

Popular Problems

$9 + 11 + 13 + 15 + \dots + 99 = \sum_{k = 0}^{45} ◻$